You are given a binary string (i. e. a string consisting of characters 0 and/or 1) s of length n. You can perform the following operation with the string s at most once: choose a substring (a contiguous subsequence) of s having exactly k characters 1 in it, and shuffle it (reorder the characters in the substring as you wish).

Calculate the number of different strings which can be obtained from s by performing this operation at most once.

Input

The first line contains two integers n and k (2 ≤ n ≤ 5000; 0 ≤ k ≤ n).

The second line contains the string s of length n, consisting of characters 0 and/or 1.

Output

Print one integer — the number of different strings which can be obtained from s by performing the described operation at most once. Since the answer can be large, output it modulo 998244353.

Examples

Input


7 2
1100110


Output


16


Input


5 0
10010


Output


1


Input


8 1
10001000


Output


10


Input


10 8
0010011000


Output


1

Note

Some strings you can obtain in the first example:

  * to obtain 0110110, you can take the substring from the 1-st character to the 4-th character, which is 1100, and reorder its characters to get 0110; 
  * to obtain 1111000, you can take the substring from the 3-rd character to the 7-th character, which is 00110, and reorder its characters to get 11000; 
  * to obtain 1100101, you can take the substring from the 5-th character to the 7-th character, which is 110, and reorder its characters to get 101. 



In the second example, k = 0 so you can only choose the substrings consisting only of 0 characters. Reordering them doesn't change the string at all, so the only string you can obtain is 10010.
5 2
00011
 5 4
11011
 81 5
100000101000000011001101010100001010111000000110000000001000101101111000001000110
 4 3
1011
 47 25
11110011111001110100101010011000110111011011101
 291 159
011001110111110110001111000110001001111100010001011000110100110010001101100110111010101111100111110000110111001100111010001000011001010011011100101000101011010110010011010100100101111010111000001011110101001011101110100100111110011001101010001110110101010101100011101110111000011111011001110
 459 15
101010101010101010101010101010101010101010101111111111111111111111010101010101010101010101011111111111111111111101010101010101010101111111111111111111111101010101010101010111111111111111111111111101010101010101111111111111111111111111110110101010101011111111111111111111111111111001000111111111111111111111111111110000011111111111111111111111111111000001111111111111111111111111111100001111111111111111111111111111100111111111111111111111111111110111111111111
 105 58
110101000111100000011011011101110100001000110011010101100010111001011111010010111101011011010110011011110
 52 11
0110011111001000010000100010010001010100000100110100

10
 5
 187538
 4
 314445058
 1
 0
 132732930
 478323746

#include<bits/stdc++.h>
using namespace std;
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class A> using ordered_set=tree<A,null_type,less<A>,rb_tree_tag,tree_order_statistics_node_update>;

#define ll long long 
#define int long long
#define pii pair<int,int>
#define vi vector<int>
#define vvi vector<vector<int>>
#define all(x) x.begin(),x.end()
#define debug(a...) cout<<#a<<": ";for(auto &it:a)cout<<it<<" ";cout<<"\n";
#define pb push_back
#define br "\n"
const int mod=998244353;
const int maxn=2e5+9;
const int INF=1e18;

// ---------------------------------------------------------------------
int n;
vector<int>sum;
// ---------------------------------------------------------------------
int getSum(int l,int r)
{
    int res=sum[r];
    if(l!=0)
        res-=sum[l-1];
    return res;
}
vector<int>fac;
void factorial()
{
    fac.assign(maxn,0);
    fac[0]=1;
    for(int i=1;i<maxn;i++)
        fac[i]=(fac[i-1]*i)%mod;
}
int binexp(int a,int b)
{
    int res=1;
    while(b!=0)
    {
        if(b%2!=0)
            res=(res*a)%mod;
        a=(a*a)%mod;
        b=b/2;
    }
    return res;
}
int ncr(int n1,int r)
{
    if(n1<r||n1==0)
        return 0;
    int den=(fac[r]*fac[n1-r])%mod;
    int res=(fac[n1]*binexp(den,mod-2))%mod;
    return res;
}
// ---------------------------------------------------------------------
void test_case(int tno)
{
    int k;
    cin>>n>>k;
    string s;
    cin>>s;
    sum.assign(n,0);
    for(int i=0;i<n;i++)
    {
        sum[i]=s[i]-'0';
        if(i!=0)
            sum[i]+=sum[i-1];
    }
    vector<int>dp(n,0);
    int flag=0;
    for(int i=0;i<n;i++)
    {
        if(i!=0)
            dp[i]=(dp[i]+dp[i-1])%mod;
        int l=-1;
        for(int j=i;j>=0;j--)
        {
            if(getSum(j,i)==k)
                l=j;
        }
        if(l==-1)
            continue;
        flag++;
        dp[i]=(dp[i]+ncr(i-l+1,k))%mod;
        dp[i]=(dp[i]-ncr(i-l,getSum(l,i-1))+2*mod)%mod;
    }
    if(flag==0||k==0)
        cout<<1<<br;
    else
        cout<<dp[n-1]<<br;
}

int32_t main()
{
    ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL);
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    #endif    
    int t=1;
    //cin>>t;
    factorial();
    for(int i=1;i<=t;i++)
        test_case(i);
}