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:
#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int maxn=3005;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;

struct Question{
	int l,r;
}q[maxn];

LL C[maxn][maxn],dp[maxn][maxn];
int sum[maxn];
char s[maxn];
int n,m;

void pre_C(){
	for(int i=0;i<=n;i++){
		C[i][0]=1;
		for(int j=1;j<=i;j++){
			C[i][j]=(C[i-1][j]+C[i-1][j-1])%MOD;
		}
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	pre_C();
	for(int i=1;i<=n;i++)sum[i]=sum[i-1]+s[i]-'0';
	for(int i=1;i<=m;i++){
		scanf("%d%d",&q[i].l,&q[i].r);
		if(q[i].r<=q[i-1].r)m--,i--;
	}
	
	q[m+1].l=n+1;q[m+1].r=n+1;
	sum[n+1]=sum[n];
	dp[1][sum[q[1].r]-sum[q[1].l-1]]=1;
	for(int i=1;i<=m;i++){
		if(q[i].r<q[i+1].l){
			int len=q[i].r-q[i].l+1;
			for(int j=0;j<=len;j++){
				dp[i+1][sum[q[i+1].r]-sum[q[i+1].l-1]]+=dp[i][j]*C[len][j]%MOD,dp[i+1][sum[q[i+1].r]-sum[q[i+1].l-1]]%=MOD;
			}
		}
		else{
			int LEN=q[i].r-q[i].l+1;//qujianchang
			int len=q[i].r-q[i+1].l+1;//jiaojichang
			int pre=q[i+1].l-q[i].l;//qianzhui
			for(int j=0;j<=LEN;j++)
				for(int k=max(0,j-pre);k<=j&&k<=len;k++){
					dp[i+1][sum[q[i+1].r]-sum[q[i].r]+k]+=dp[i][j]*C[pre][j-k]%MOD,dp[i+1][sum[q[i+1].r]-sum[q[i].r]+k]%=MOD;
				}
		}
	}
	printf("%lld\n",dp[m+1][0]);
	return 0;
}