This is the easy version of the problem. The difference between the versions is that the easy version does not require you to output the numbers of the rods to be removed. You can make hacks only if all versions of the problem are solved.

Stitch likes experimenting with different machines with his friend Sparky. Today they built another machine.

The main element of this machine are n rods arranged along one straight line and numbered from 1 to n inclusive. Each of these rods must carry an electric charge quantitatively equal to either 1 or -1 (otherwise the machine will not work). Another condition for this machine to work is that the sign-variable sum of the charge on all rods must be zero.

More formally, the rods can be represented as an array of n numbers characterizing the charge: either 1 or -1. Then the condition must hold: a_1 - a_2 + a_3 - a_4 + … = 0, or ∑_{i=1}^n (-1)^{i-1} ⋅ a_i = 0.

Sparky charged all n rods with an electric current, but unfortunately it happened that the rods were not charged correctly (the sign-variable sum of the charge is not zero). The friends decided to leave only some of the rods in the machine. Sparky has q questions. In the ith question Sparky asks: if the machine consisted only of rods with numbers l_i to r_i inclusive, what minimal number of rods could be removed from the machine so that the sign-variable sum of charges on the remaining ones would be zero? Perhaps the friends got something wrong, and the sign-variable sum is already zero. In that case, you don't have to remove the rods at all.

If the number of rods is zero, we will assume that the sign-variable sum of charges is zero, that is, we can always remove all rods.

Help your friends and answer all of Sparky's questions!

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1 ≤ t ≤ 10^3), denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains two positive integers n and q (1 ≤ n, q ≤ 3 ⋅ 10^5) — the number of rods and the number of questions.

The second line of each test case contains a non-empty string s of length n, where the charge of the i-th rod is 1 if s_i is the "+" symbol, or -1 if s_i is the "-" symbol.

Each next line from the next q lines contains two positive integers l_i ans r_i (1 ≤ l_i ≤ r_i ≤ n) — numbers, describing Sparky's questions.

It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5, and the sum of q over all test cases does not exceed 3 ⋅ 10^5.

Output

For each test case, print a single integer — the minimal number of rods that can be removed.

Example

Input


3
14 1
+--++---++-++-
1 14
14 3
+--++---+++---
1 14
6 12
3 10
4 10
+-+-
1 1
1 2
1 3
1 4
2 2
2 3
2 4
3 3
3 4
4 4


Output


2
2
1
0
1
2
1
2
1
2
1
1
2
1

Note

In the first test case for the first query you can remove the rods numbered 5 and 8, then the following set of rods will remain: +--+--++-++-. It is easy to see that here the sign-variable sum is zero.

In the second test case:

  * For the first query, we can remove the rods numbered 1 and 11, then the following set of rods will remain: --++---++---. It is easy to see that here the sign-variable sum is zero. 
  * For the second query we can remove the rod numbered 9, then the following set of rods will remain: ---++-. It is easy to see that here the variable sum is zero. 
  * For the third query we can not remove the rods at all. 

def divisors(M):
    d=[]
    i=1
    while M>=i**2:
        if M%i==0:
            d.append(i)
            if i**2!=M:
                d.append(M//i)
        i=i+1
    return d

def popcount(x):
    x = x - ((x >> 1) & 0x55555555)
    x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
    x = (x + (x >> 4)) & 0x0f0f0f0f
    x = x + (x >> 8)
    x = x + (x >> 16)
    return x & 0x0000007f

def eratosthenes(n):
    res=[0 for i in range(n+1)]
    prime=set([])
    for i in range(2,n+1):
        if not res[i]:
            prime.add(i)
            for j in range(1,n//i+1):
                res[i*j]=1
    return prime

def factorization(n):
    res=[]
    for p in prime:
        if n%p==0:
            while n%p==0:
                n//=p
            res.append(p)
    if n!=1:
        res.append(n)
    return res

def euler_phi(n):
    res = n
    for x in range(2,n+1):
        if x ** 2 > n:
            break
        if n%x==0:
            res = res//x * (x-1)
            while n%x==0:
                n //= x
    if n!=1:
        res = res//n * (n-1)
    return res

def ind(b,n):
    res=0
    while n%b==0:
        res+=1
        n//=b
    return res

def isPrimeMR(n):
    d = n - 1
    d = d // (d & -d)
    L = [2, 3, 5, 7, 11, 13, 17]
    for a in L:
        t = d
        y = pow(a, t, n)
        if y == 1: continue
        while y != n - 1:
            y = (y * y) % n
            if y == 1 or t == n - 1: return 0
            t <<= 1
    return 1
def findFactorRho(n):
    from math import gcd
    m = 1 << n.bit_length() // 8
    for c in range(1, 99):
        f = lambda x: (x * x + c) % n
        y, r, q, g = 2, 1, 1, 1
        while g == 1:
            x = y
            for i in range(r):
                y = f(y)
            k = 0
            while k < r and g == 1:
                ys = y
                for i in range(min(m, r - k)):
                    y = f(y)
                    q = q * abs(x - y) % n
                g = gcd(q, n)
                k += m
            r <<= 1
        if g == n:
            g = 1
            while g == 1:
                ys = f(ys)
                g = gcd(abs(x - ys), n)
        if g < n:
            if isPrimeMR(g): return g
            elif isPrimeMR(n // g): return n // g
            return findFactorRho(g)
def primeFactor(n):
    i = 2
    ret = {}
    rhoFlg = 0
    while i*i <= n:
        k = 0
        while n % i == 0:
            n //= i
            k += 1
        if k: ret[i] = k
        i += 1 + i % 2
        if i == 101 and n >= 2 ** 20:
            while n > 1:
                if isPrimeMR(n):
                    ret[n], n = 1, 1
                else:
                    rhoFlg = 1
                    j = findFactorRho(n)
                    k = 0
                    while n % j == 0:
                        n //= j
                        k += 1
                    ret[j] = k

    if n > 1: ret[n] = 1
    if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
    return ret

def divisors(n):
    res = [1]
    prime = primeFactor(n)
    for p in prime:
        newres = []
        for d in res:
            for j in range(prime[p]+1):
                newres.append(d*p**j)
        res = newres
    res.sort()
    return res

def xorfactorial(num):
    if num==0:
        return 0
    elif num==1:
        return 1
    elif num==2:
        return 3
    elif num==3:
        return 0
    else:
        x=baseorder(num)
        return (2**x)*((num-2**x+1)%2)+function(num-2**x)

def xorconv(n,X,Y):
    if n==0:
        res=[(X[0]*Y[0])%mod]
        return res
    x=[X[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
    y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
    z=[X[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
    w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
    res1=xorconv(n-1,x,y)
    res2=xorconv(n-1,z,w)
    former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
    latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
    former=list(map(lambda x:x%mod,former))
    latter=list(map(lambda x:x%mod,latter))
    return former+latter

def merge_sort(A,B):
    pos_A,pos_B = 0,0
    n,m = len(A),len(B)
    res = []
    while pos_A < n and pos_B < m:
        a,b = A[pos_A],B[pos_B]
        if a < b:
            res.append(a)
            pos_A += 1
        else:
            res.append(b)
            pos_B += 1
    res += A[pos_A:]
    res += B[pos_B:]
    return res

class UnionFindVerSize():
    def __init__(self, N):
        self._parent = [n for n in range(0, N)]
        self._size = [1] * N
        self.group = N

    def find_root(self, x):
        if self._parent[x] == x: return x
        self._parent[x] = self.find_root(self._parent[x])
        stack = [x]
        while self._parent[stack[-1]]!=stack[-1]:
            stack.append(self._parent[stack[-1]])
        for v in stack:
            self._parent[v] = stack[-1]
        return self._parent[x]

    def unite(self, x, y):
        gx = self.find_root(x)
        gy = self.find_root(y)
        if gx == gy: return

        self.group -= 1

        if self._size[gx] < self._size[gy]:
            self._parent[gx] = gy
            self._size[gy] += self._size[gx]
        else:
            self._parent[gy] = gx
            self._size[gx] += self._size[gy]

    def get_size(self, x):
        return self._size[self.find_root(x)]

    def is_same_group(self, x, y):
        return self.find_root(x) == self.find_root(y)

class WeightedUnionFind():
    def __init__(self,N):
        self.parent = [i for i in range(N)]
        self.size = [1 for i in range(N)]
        self.val = [0 for i in range(N)]
        self.flag = True
        self.edge = [[] for i in range(N)]

    def dfs(self,v,pv):
        stack = [(v,pv)]
        new_parent = self.parent[pv]
        while stack:
            v,pv = stack.pop()
            self.parent[v] = new_parent
            for nv,w in self.edge[v]:
                if nv!=pv:
                    self.val[nv] = self.val[v] + w
                    stack.append((nv,v))

    def unite(self,x,y,w):
        if not self.flag:
            return
        if self.parent[x]==self.parent[y]:
            self.flag = (self.val[x] - self.val[y] == w)
            return

        if self.size[self.parent[x]]>self.size[self.parent[y]]:
            self.edge[x].append((y,-w))
            self.edge[y].append((x,w))
            self.size[x] += self.size[y]
            self.val[y] = self.val[x] - w
            self.dfs(y,x)
        else:
            self.edge[x].append((y,-w))
            self.edge[y].append((x,w))
            self.size[y] += self.size[x]
            self.val[x] = self.val[y] + w
            self.dfs(x,y)

class Dijkstra():
    class Edge():
        def __init__(self, _to, _cost):
            self.to = _to
            self.cost = _cost

    def __init__(self, V):
        self.G = [[] for i in range(V)]
        self._E = 0
        self._V = V

    @property
    def E(self):
        return self._E

    @property
    def V(self):
        return self._V

    def add_edge(self, _from, _to, _cost):
        self.G[_from].append(self.Edge(_to, _cost))
        self._E += 1

    def shortest_path(self, s):
        import heapq
        que = []
        d = [10**15] * self.V
        d[s] = 0
        heapq.heappush(que, (0, s))

        while len(que) != 0:
            cost, v = heapq.heappop(que)
            if d[v] < cost: continue

            for i in range(len(self.G[v])):
                e = self.G[v][i]
                if d[e.to] > d[v] + e.cost:
                    d[e.to] = d[v] + e.cost
                    heapq.heappush(que, (d[e.to], e.to))
        return d

#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
    N = len(s)
    Z_alg = [0]*N

    Z_alg[0] = N
    i = 1
    j = 0
    while i < N:
        while i+j < N and s[j] == s[i+j]:
            j += 1
        Z_alg[i] = j
        if j == 0:
            i += 1
            continue
        k = 1
        while i+k < N and k + Z_alg[k]<j:
            Z_alg[i+k] = Z_alg[k]
            k += 1
        i += k
        j -= k
    return Z_alg

class BIT():
    def __init__(self,n,mod=0):
        self.BIT = [0]*(n+1)
        self.num = n
        self.mod = mod

    def query(self,idx):
        res_sum = 0
        mod = self.mod
        while idx > 0:
            res_sum += self.BIT[idx]
            if mod:
                res_sum %= mod
            idx -= idx&(-idx)
        return res_sum

    #Ai += x O(logN)
    def update(self,idx,x):
        mod = self.mod
        while idx <= self.num:
            self.BIT[idx] += x
            if mod:
                self.BIT[idx] %= mod
            idx += idx&(-idx)
        return

class dancinglink():
    def __init__(self,n,debug=False):
        self.n = n
        self.debug = debug
        self._left = [i-1 for i in range(n)]
        self._right = [i+1 for i in range(n)]
        self.exist = [True for i in range(n)]

    def pop(self,k):
        if self.debug:
            assert self.exist[k]
        L = self._left[k]
        R = self._right[k]
        if L!=-1:
            if R!=self.n:
                self._right[L],self._left[R] = R,L
            else:
                self._right[L] = self.n
        elif R!=self.n:
            self._left[R] = -1
        self.exist[k] = False

    def left(self,idx,k=1):
        if self.debug:
            assert self.exist[idx]
        res = idx
        while k:
            res = self._left[res]
            if res==-1:
                break
            k -= 1
        return res

    def right(self,idx,k=1):
        if self.debug:
            assert self.exist[idx]
        res = idx
        while k:
            res = self._right[res]
            if res==self.n:
                break
            k -= 1
        return res

class SparseTable():
    def __init__(self,A,merge_func,ide_ele):
        N=len(A)
        n=N.bit_length()
        self.table=[[ide_ele for i in range(n)] for i in range(N)]
        self.merge_func=merge_func

        for i in range(N):
            self.table[i][0]=A[i]

        for j in range(1,n):
            for i in range(0,N-2**j+1):
                f=self.table[i][j-1]
                s=self.table[i+2**(j-1)][j-1]
                self.table[i][j]=self.merge_func(f,s)

    def query(self,s,t):
        b=t-s+1
        m=b.bit_length()-1
        return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])

class BinaryTrie:
    class node:
        def __init__(self,val):
            self.left = None
            self.right = None
            self.max = val

    def __init__(self):
        self.root = self.node(-10**15)

    def append(self,key,val):
        pos = self.root
        for i in range(29,-1,-1):
            pos.max = max(pos.max,val)
            if key>>i & 1:
                if pos.right is None:
                    pos.right = self.node(val)
                    pos = pos.right
                else:
                    pos = pos.right
            else:
                if pos.left is None:
                    pos.left = self.node(val)
                    pos = pos.left
                else:
                    pos = pos.left
        pos.max = max(pos.max,val)

    def search(self,M,xor):
        res = -10**15
        pos = self.root
        for i in range(29,-1,-1):
            if pos is None:
                break

            if M>>i & 1:
                if xor>>i & 1:
                    if pos.right:
                        res = max(res,pos.right.max)
                    pos = pos.left
                else:
                    if pos.left:
                        res = max(res,pos.left.max)
                    pos = pos.right
            else:
                if xor>>i & 1:
                    pos = pos.right
                else:
                    pos = pos.left

        if pos:
            res = max(res,pos.max)
        return res

def solveequation(edge,ans,n,m):
    #edge=[[to,dire,id]...]
    x=[0]*m
    used=[False]*n
    for v in range(n):
        if used[v]:
            continue
        y = dfs(v)
        if y!=0:
            return False
    return x

    def dfs(v):
        used[v]=True
        r=ans[v]
        for to,dire,id in edge[v]:
            if used[to]:
                continue
            y=dfs(to)
            if dire==-1:
                x[id]=y
            else:
                x[id]=-y
            r+=y
        return r

class SegmentTree:
    def __init__(self, init_val, segfunc, ide_ele):
        n = len(init_val)
        self.segfunc = segfunc
        self.ide_ele = ide_ele
        self.num = 1 << (n - 1).bit_length()
        self.tree = [ide_ele] * 2 * self.num
        self.size = n
        for i in range(n):
            self.tree[self.num + i] = init_val[i]
        for i in range(self.num - 1, 0, -1):
            self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])

    def update(self, k, x):
        k += self.num
        self.tree[k] = x
        while k > 1:
            self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
            k >>= 1

    def query(self, l, r):
        if r==self.size:
            r = self.num

        res = self.ide_ele

        l += self.num
        r += self.num
        while l < r:
            if l & 1:
                res = self.segfunc(res, self.tree[l])
                l += 1
            if r & 1:
                res = self.segfunc(res, self.tree[r - 1])
            l >>= 1
            r >>= 1
        return res

    def bisect_l(self,l,r,x):
        l += self.num
        r += self.num
        Lmin = -1
        Rmin = -1
        while l<r:
            if l & 1:
                if self.tree[l] <= x and Lmin==-1:
                    Lmin = l
                l += 1
            if r & 1:
                if self.tree[r-1] <=x:
                    Rmin = r-1
            l >>= 1
            r >>= 1

        if Lmin != -1:
            pos = Lmin
            while pos<self.num:
                if self.tree[2 * pos] <=x:
                    pos = 2 * pos
                else:
                    pos = 2 * pos +1
            return pos-self.num
        elif Rmin != -1:
            pos = Rmin
            while pos<self.num:
                if self.tree[2 * pos] <=x:
                    pos = 2 * pos
                else:
                    pos = 2 * pos +1
            return pos-self.num
        else:
            return -1

import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd,log

input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())

c = {"+":1,"-":-1}

for _ in range(int(input())):
    n,q = mi()
    S = input()
    A = [0]
    for i in range(n):
        if i&1:
            A.append(-c[S[i]])
        else:
            A.append(c[S[i]])
    for i in range(1,n+1):
        A[i] += A[i-1]
    
    for i in range(q):
        l,r = mi()
        if (r-l+1)&1:
            print(1)
        else:
            tmp = A[r] - A[l-1]
            if tmp==0:
                print(0)
            else:
                print(2)
