Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

5 2
01001
2 4
3 5

Sample Output 1

6

After the first operation, S can be one of the following three: 01001, 00101 and 00011.

After the second operation, S can be one of the following six: 01100, 01010, 01001, 00011, 00101 and 00110.

Sample Input 2

9 3
110111110
1 4
4 6
6 9

Sample Output 2

26

Sample Input 3

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

Sample Output 3

143

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#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[2][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)lol(j,2)d[j][i]=0;d[0][0]=1;
    for(ll i=0,rnd=0,bef=0;i<=n;i++){
	ll cnt=0;
	while(!Q.empty()){
	    if(not(Q.front().X<=i))break;
	    for(ll j=bef;j<rnd;j++)cnt+=s[j]-'0';
	    bef=rnd;Q.pop();rnd=Q.front().Y;
	}
	lol(j,N)d[1^i&1][j]=0;
	ll lim=max(0LL,bef-i);
	for(ll j=cnt;j<=lim;j++)Mad(d[1^i&1][j],d[i&1][j-cnt]);
	for(ll j=cnt;j<=lim;j++)Mad(d[1^i&1][j],d[i&1][j-cnt+1]);
    }
    cout<<d[1^n&1][0]<<endl;
    return 0;
}