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 <cctype>
#include <algorithm>
#define gc() getchar()
#define mod 1000000007
#define inv2 500000004ll
typedef long long LL;
const int N=3005;

inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-48,c=gc());
	return now;
}
inline int FP(int x,int k)
{
	int t=1;
	for(; k; k>>=1,x=1ll*x*x%mod)
		if(k&1) t=1ll*t*x%mod;
	return t;
}

int main()
{
	static int A[N],f[N][N];
	const int n=read(),q=read();
	for(int i=1; i<=n; ++i) A[i]=read();
	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=read(),y=read();
		f[x][y]=f[y][x]=inv2*(f[x][y]+f[y][x])%mod;
		for(int j=1; j<=n; ++j)
			if(j!=x && j!=y)
				f[j][x]=f[j][y]=inv2*(f[j][x]+f[j][y])%mod,
				f[x][j]=f[y][j]=inv2*(f[x][j]+f[y][j])%mod;
	}
	LL ans=0;
	for(int i=1; i<=n; ++i)
		for(int j=1; j<i; ++j) ans+=f[i][j];
	printf("%lld\n",ans%mod*FP(2,q)%mod);

	return 0;
}