Problem: You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
struct ee{int go,next;} e[200005];
int n,q,b[200005],s[200005],ans,cnt,f[200005],mfs[800005],mf[800005],ljb[200005];
void build(int p,int ll,int rr){
	if (ll==rr){
		if (!ll) mf[p]=mfs[p]=0;
		else mf[p]=mfs[p]=-1000000000;
		return;
	}
	build(p*2,ll,(ll+rr)/2);build(p*2+1,(ll+rr)/2+1,rr);
	mfs[p]=max(mfs[p*2],mfs[p*2+1]);
}
int askf(int p,int ll,int rr,int x,int y){
	if (x>y) return 0;
	if (ll==x&&rr==y) return mf[p];
	int mid=(ll+rr)/2;
	if (y<=mid) return askf(p*2,ll,mid,x,y);
	else if (x>mid) return askf(p*2+1,mid+1,rr,x,y);
	else return max(askf(p*2,ll,mid,x,mid),askf(p*2+1,mid+1,rr,mid+1,y));
}
int askfs(int p,int ll,int rr,int x,int y){
	if (ll==x&&rr==y) return mfs[p];
	int mid=(ll+rr)/2;
	if (y<=mid) return askfs(p*2,ll,mid,x,y);
	else if (x>mid) return askfs(p*2+1,mid+1,rr,x,y);
	else return max(askfs(p*2,ll,mid,x,mid),askfs(p*2+1,mid+1,rr,mid+1,y));
}
void add(int p,int ll,int rr,int x,int nf,int nfs){
	mf[p]=max(mf[p],nf);mfs[p]=max(mfs[p],nfs);
	if (ll==rr) return;
	int mid=(ll+rr)/2;
	if (x<=mid) add(p*2,ll,mid,x,nf,nfs);
	else add(p*2+1,mid+1,rr,x,nf,nfs);
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&b[i]);
	for (int i=1;i<=n;i++) s[i]=s[i-1]+(b[i]?1:-1),ans+=!b[i]; 
	scanf("%d",&q);
	for (int i=1;i<=q;i++){
		int x,y;scanf("%d%d",&x,&y);
		e[++cnt]=(ee){y,ljb[x]};ljb[x]=cnt;
	}
	build(1,0,n);f[0]=0;
	for (int i=1;i<=n;i++) f[i]=-1000000000;
	for (int i=1;i<=n;i++){
		for (int td=ljb[i];td;td=e[td].next){
			int j=e[td].go;
			f[j]=max(f[j],askf(1,0,n,0,i-1)+s[j]-s[i-1]);
			if (i<j) f[j]=max(f[j],askfs(1,0,n,i,j-1)+s[j]);
		}
		for (int td=ljb[i];td;td=e[td].next){
			int j=e[td].go;
			add(1,0,n,j,f[j],f[j]-s[j]);
		}
	}
	printf("%d",n-ans-askf(1,0,n,1,n));
}