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;
#define MOD 1000000007
#define LL long long
#define N 3011
int n,q,a[N],x,y;
LL f[N][N],ans,inv2; 
LL q_pow(LL a,LL b){
	LL res=1;
	while(b){
		if(b&1ll) res=res*a%MOD;
		b>>=1ll;
		a=a*a%MOD;
	}
	return res;
}
int main(){
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
	  for(int j=1;j<=n;j++){
	  	if(i==j) continue;
	  	f[i][j]=(a[i]>a[j]);
	  }
	inv2=q_pow(2,MOD-2);
	for(int Q=1;Q<=q;Q++){
		scanf("%d%d",&x,&y);
		swap(a[x],a[y]); 
		for(int i=1;i<=n;i++){
			if(i!=x&&i!=y){
				f[x][i]=f[y][i]=inv2*(f[x][i]+f[y][i])%MOD;
				f[i][x]=f[i][y]=inv2*(f[i][x]+f[i][y])%MOD;
			}
		}
		f[x][y]=f[y][x]=inv2*(f[x][y]+f[y][x])%MOD;
	}
	ans=0;
	for(int i=1;i<=n;i++)
	  for(int j=1;j<i;j++)
	    ans=(ans+f[j][i])%MOD;
	ans=ans*q_pow(2,q)%MOD;
	printf("%lld\n",ans);
}