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 signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

int N,M;
string S;
ll mo=1000000007;
ll dp[3030][3030];

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>M>>S;
	
	int CR=0,CL=0;
	int B[2]={};
	dp[0][0]=1;
	while(M--) {
		int L,R;
		cin>>L>>R;
		L--,R--;
		while(CL<L) {
			if(CL==CR) B[S[CR++]=='1']++;
			FOR(i,3001) if(dp[CL][i]) {
				if(i<B[0]) (dp[CL+1][i+1]+=dp[CL][i])%=mo;
				if(CL-i<B[1]) (dp[CL+1][i]+=dp[CL][i])%=mo;
			}
			CL++;
		}
		while(CR<=R) B[S[CR++]=='1']++;
	}
	while(CL<N) {
		if(CL==CR) B[S[CR++]=='1']++;
		FOR(i,3001) if(dp[CL][i]) {
			if(i<B[0]) (dp[CL+1][i+1]+=dp[CL][i])%=mo;
			if(CL-i<B[1]) (dp[CL+1][i]+=dp[CL][i])%=mo;
		}
		CL++;
	}
	
	cout<<dp[CL][B[0]]<<endl;
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}