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>
#define cs const
using namespace std;
cs int Mod = 1e9 + 7, iv2 = (Mod+1)>>1;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
void Add(int &a, int b){ a = add(a,b); }
void Dec(int &a, int b){ a = dec(a,b); }
void Mul(int &a, int b){ a = mul(a,b); }
cs int N = 3e3 + 50;
int n, m, a[N], dp[N][N];
int main(){
//	freopen("1.in","r",stdin);
	scanf("%d%d",&n,&m);
	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(a[i]<a[j]) dp[i][j]=1; int coe=1;
	for(int i=1,x,y; i<=m; i++){
		scanf("%d%d",&x,&y); Mul(coe,2);
		dp[x][y]=dp[y][x]=mul(iv2,add(dp[x][y],dp[y][x]));
		for(int j=1; j<=n; j++) if(j!=x&&j!=y){
			dp[x][j]=dp[y][j]=mul(iv2,add(dp[x][j],dp[y][j]));
			dp[j][x]=dp[j][y]=mul(iv2,add(dp[j][x],dp[j][y]));
		} 
	} 
	int as = 0;
	for(int i=1; i<=n; i++)
	for(int j=1; j<i; j++) Add(as,dp[i][j]);
	cout<<mul(as,coe); return 0;
}