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
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
#define pb push
#define mp make_pair
#define X first
#define Y second
#define Mad(a,b) a=(a+b)%mod;
typedef long long ll;

using namespace std;
typedef pair<ll,ll> P;
queue<P> Q;
#define N 3010
ll n,m,d[N];string s;
int main(){
    cin>>n>>m>>s;
    Q.pb(mp(0,0));
    while(m--){
	ll a,b;cin>>a>>b;
	if(Q.back().Y<b)Q.pb(mp(a,b));
    }
    lol(i,N)d[i]=0;d[0]=1;
    for(ll i=0,rnd=0;i<=n;i++){
	ll cnt=0;
	while(!Q.empty()){
	    if(not(Q.front().X<=i))break;
	    for(ll j=max(i-1,rnd);j<Q.front().Y;j++)cnt+=s[j]-'0';
	    rnd=Q.front().Y;Q.pop();
	}
	for(ll j=N-1;j>=0;j--)d[j]=(j-cnt>=0?d[j-cnt]:0);
	ll lim=max(0LL,rnd-i);
	for(ll j=0;j<=lim;j++)Mad(d[j],d[j+1]);
	for(ll j=lim+1;j<N;j++)d[j]=0;
    }
    cout<<d[0]<<endl;
    return 0;
}
