Greg has a pad. The pad's screen is an n × m rectangle, each cell can be either black or white. We'll consider the pad rows to be numbered with integers from 1 to n from top to bottom. Similarly, the pad's columns are numbered with integers from 1 to m from left to right.

Greg thinks that the pad's screen displays a cave if the following conditions hold:

  * There is a segment [l, r] (1 ≤ l ≤ r ≤ n), such that each of the rows l, l + 1, ..., r has exactly two black cells and all other rows have only white cells. 
  * There is a row number t (l ≤ t ≤ r), such that for all pairs of rows with numbers i and j (l ≤ i ≤ j ≤ t) the set of columns between the black cells in row i (with the columns where is these black cells) is the subset of the set of columns between the black cells in row j (with the columns where is these black cells). Similarly, for all pairs of rows with numbers i and j (t ≤ i ≤ j ≤ r) the set of columns between the black cells in row j (with the columns where is these black cells) is the subset of the set of columns between the black cells in row i (with the columns where is these black cells). 



Greg wondered, how many ways there are to paint a cave on his pad. Two ways can be considered distinct if there is a cell that has distinct colors on the two pictures.

Help Greg.

Input

The first line contains two integers n, m — the pad's screen size (1 ≤ n, m ≤ 2000).

Output

In the single line print the remainder after dividing the answer to the problem by 1000000007 (109 + 7).

Examples

Input

1 1


Output

0


Input

4 4


Output

485


Input

3 5


Output

451
#pragma GCC optimize("Ofast")

#include <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <ratio>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
 
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
 
 
#define FLSH fflush(stdout)
#define fileIO(name) \
    freopen(name".in", "r", stdin); \
    freopen(name".out", "w", stdout);
#define PRECISION(x) cout << fixed << setprecision(x); 
#define FAST_IO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
 
 
typedef tree<int,null_type,less<int>,rb_tree_tag,
tree_order_statistics_node_update> indexed_set;
 
typedef long long ll;
typedef long double LD;
#define int ll
#define double LD
#define pb push_back
#define mp make_pair
#define REP(i,n) for (int i = 0; i < n; i++)
#define REP1(i,n) for (int i = 1; i<=n; i++)
#define FOR(i,a,b) for (int i = a; i < b; i++)
#define REPD(i,n) for (int i = n-1; i >= 0; i--)
#define REPD1(i,n) for (int i = n; i > 0; i--)
#define FORD(i,a,b) for (int i = a; i >= b; i--)
#define foreach(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define remax(a,b) a = max(a,b)
#define remin(a,b) a = min(a,b)
#define all(v) v.begin(),v.end()
typedef map<int,int> mii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;
#define F first
#define S second
#define PQ(type) priority_queue<type>
#define PQD(type) priority_queue<type,vector<type>,greater<type> >
#define ITR :: iterator it
#define WL(t) while(t --)
#define SZ(x) ((int)(x).size())
#define runtime() ((double)clock() / CLOCKS_PER_SEC)
#define TR(container,it) for(typeof(container.begin()) it=container.begin();it!=container.end();it++)
#define sqr(x) ((x)*(x))
 
const int MAXN = 1000005;
const int SQRTN = 1003;
const int LOGN = 22;
const double PI=acos(-1);
const int INF = 1000000000;
const int MOD = 1000000007;
const int FMOD = 998244353;
const double eps = 1e-9;
 
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count()); 
#define SHUF(v) shuffle(all(v), RNG);
 
template<typename T> T gcd(T a, T b){return(b?__gcd(a,b):a);}
template<typename T> T lcm(T a, T b){return(a*(b/gcd(a,b)));}
int add(int a, int b, int c){int res=a+b;return(res>=c?res-c:res);}
int mod_neg(int a, int b, int c){int res;if(abs(a-b)<c)res=a-b;else res=(a-b)%c;return(res<0?res+c:res);}
int mul(int a, int b, int c){ll res=(ll)a*b;return(res>=c?res%c:res);}
ll mulmod(ll a,ll b, ll m){ll q = (ll)(((LD)a*(LD)b)/(LD)m);ll r=a*b-q*m;if(r>m)r%=m;if(r<0)r+=m;return r;}
template<typename T>T expo(T e, T n){T x=1,p=e;while(n){if(n&1)x=x*p;p=p*p;n>>=1;}return x;}
template<typename T>T power(T e, T n, T m){T x=1,p=e;while(n){if(n&1)x=mul(x,p,m);p=mul(p,p,m);n>>=1;}return x;}
template<typename T>T extended_euclid(T a, T b, T &x, T &y){T xx=0,yy=1;y=0;x=1;while(b){T q=a/b,t=b;b=a%b;a=t;\
t=xx;xx=x-q*xx;x=t;t=yy;yy=y-q*yy;y=t;}return a;}
template<typename T>T mod_inverse(T a, T n){T x,y,z=0;T d=extended_euclid(a,n,x,y);return(d>1?-1:mod_neg(x,z,n));}
 
using namespace std;

const int mx = 1e5+10;
int n, m, k;
int ori[mx], le[mx], re[mx], ad[mx];
int t[4*mx];

void addseg(int l, int r, int v, int s = 0, int e = n, int x = 1){
	if(l >= e || r <= s) return;
	if(l <= s && e <= r){
		t[x] += v;
		return;
	}
	int mid = (e + s) / 2;
	addseg(l, r, v, s, mid, 2*x);
	addseg(l, r, v, mid, e, 2*x+1);
}

int get(int f, int s = 0, int e = n, int x = 1){
	if(e == s+1) return t[x];
	int mid = (e + s) / 2;
	if(mid > f) return t[x] + get(f, s, mid, 2*x);
	else return t[x] + get(f, mid, e, 2*x+1);
}

signed main(){
	FAST_IO
	PRECISION(10)
	cin >> n >> m >> k;
	REP(i, n) cin >> ori[i];
	REP(i, m) cin >> le[i] >> re[i] >> ad[i], le[i]--;
	vi pre(n+1);
	REP1(i, k){
		int x, y;
		cin >> x >> y;
		--x;
		pre[x]++;
		pre[y]--;
	}
	REP(i, m){
		if(i) pre[i]+=pre[i-1];
		ad[i]*=pre[i];
		addseg(le[i], re[i], ad[i]);
	}
	REP(i, n) cout << ori[i] + get(i) << " ";
	cout << endl;
}
