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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int gi(){
	int x=0,w=1;char ch=getchar();
	while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if (ch=='-') w=0,ch=getchar();
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
const int N = 3005;
const int mod = 1000000007;
int n,m,a[N],f[N][N],ans;
int main(){
	n=gi();m=gi();
	for (int i=1;i<=n;++i) a[i]=gi();
	for (int i=1;i<=n;++i)
		for (int j=1;j<=n;++j)
			f[i][j]=(a[i]<a[j]);
	for (int t=1;t<=m;++t){
		int x=gi(),y=gi();
		for (int i=1;i<=n;++i)
			if (i!=x&&i!=y){
				f[i][x]=f[i][y]=500000004ll*(f[i][x]+f[i][y])%mod;
				f[x][i]=f[y][i]=500000004ll*(f[x][i]+f[y][i])%mod;
			}
		f[x][y]=f[y][x]=500000004ll*(f[x][y]+f[y][x])%mod;
	}
	for (int i=1;i<=n;++i)
		for (int j=i+1;j<=n;++j)
			ans=(ans+f[j][i])%mod;
	for (int i=1;i<=m;++i) ans=(ans<<1)%mod;
	printf("%d\n",ans);
	return 0;
}