Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int mod=1000000007,inv_2=(mod+1)/2;
typedef long long ll;
const int N=3010;
int f[N][N],g[N][N],a[N],b[N],n,q,pw=1,ans;

int main(){
#ifdef ONLINE_JUDGE
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#endif
	cin>>n>>q;
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			f[i][j]=(a[i]<a[j]);
	for(int i=1;i<=q;++i){
		int x,y;cin>>x>>y;
		for(int j=1;j<=n;++j)b[j]=j;swap(b[x],b[y]);
		for(int j=1;j<=n;++j){
			g[x][j]=(ll)(f[b[x]][b[j]]+f[x][j])*inv_2%mod;
			g[j][x]=(ll)(f[b[j]][b[x]]+f[j][x])*inv_2%mod;
			g[y][j]=(ll)(f[b[y]][b[j]]+f[y][j])*inv_2%mod;
			g[j][y]=(ll)(f[b[j]][b[y]]+f[j][y])*inv_2%mod;
		}
		for(int j=1;j<=n;++j){
			f[x][j]=g[x][j];
			f[j][x]=g[j][x];
			f[y][j]=g[y][j];
			f[j][y]=g[j][y];
		}
		pw=(pw+pw)%mod;
	}
	for(int i=1;i<=n;++i)
		for(int j=1;j<i;++j)
			ans=(ans+(ll)pw*f[i][j])%mod;
	cout<<ans<<'\n';
	return 0;
}