Problem: There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

Solution (in CPP):

Solution:
//Link : https://codeforces.com/contest/1279/problem/D

#include <bits/stdc++.h>

using namespace std;
#define ll long long

#define N 3002
#define mod 1000000007
int add(int x,int y) {
  return (x+y)%mod;
}
int mul(int x,int y) {
  return (ll)x*y%mod;
}
int dp[N][N];
int maxR[N];
char s[N];
int preNumZero[N];
int solveDp(int ind,int zero,int n) {
  if(ind>n) {
    assert(zero==0);
    return 1;
  }
  int &ret = dp[ind][zero];
  if(ret!=-1) {
    return ret;
  }
  if(maxR[ind-1]<maxR[ind]) {
    zero += preNumZero[maxR[ind]] - preNumZero[maxR[ind-1]];
  }
  int one = (maxR[ind] - preNumZero[maxR[ind]])
    - ( ind-1 - preNumZero[maxR[ind]] +zero);
  ret = 0;
  if(zero>0) {
    ret = add(ret, solveDp(ind+1,zero-1,n));
  }
  if(one>0) {
    ret = add(ret, solveDp(ind+1,zero,n));
  }
  return ret;
}
void solve() {
  int n,m;scanf("%d %d ", &n,&m);
  scanf(" %s ", s+1);
  for(int i=1;i<=n;++i) {
    preNumZero[i] = preNumZero[i-1] + (s[i]=='0' ? 1 : 0);
  }
  for(int i=0;i<m;++i) {
    int l,r;scanf("%d %d ", &l,&r);
    maxR[l] = max(maxR[l],r);
  }
  for(int i=1;i<=n;++i) {
    maxR[i] = max(maxR[i],maxR[i-1]);
    maxR[i] = max(maxR[i],i);
  }
  memset(dp,-1,sizeof(dp));
  printf("%d\n", solveDp(1,0,n));
}
int main() {
    //freopen("input.txt","r",stdin);
    solve();
    return 0;
}