#include <bits/stdc++.h>
using namespace std;
const int maxn=200010;
const int INF=(1<<30);
int n,m,T[maxn*4],b[maxn],s[maxn],dp[maxn],ans;
vector<int> g[maxn];
void build(int l,int r,int root) {
	T[root]=-(1<<30);
	if (l==r) return;
	int mid=(l+r)/2;
	build(l,mid,root*2);
	build(mid+1,r,root*2+1);
}
int query(int L,int R,int l,int r,int root) {
	if (L<=l&&r<=R) return T[root];
	int mid=(l+r)/2,res=-INF;
	if (L<=mid) res=max(res,query(L,R,l,mid,root*2));
	if (mid<R) res=max(res,query(L,R,mid+1,r,root*2+1));
	return res;
}
void update(int pos,int l,int r,int root,int val) {
	if (l==r) {
		T[root]=max(T[root],val); return;
	}
	int mid=(l+r)/2;
	if (pos<=mid) update(pos,l,mid,root*2,val);
	else update(pos,mid+1,r,root*2+1,val);
	T[root]=max(T[root*2],T[root*2+1]);
}
int main() {
	//freopen("1.txt","r",stdin);
	scanf("%d",&n);
	for (int i=1;i<=n;i++) {
		scanf("%d",&b[i]);
		s[i]=s[i-1]+b[i];
	}
	scanf("%d",&m);
	for (int i=1;i<=m;i++) {
		int l,r; scanf("%d %d",&l,&r);
		g[l].push_back(r);
	}
	build(0,n,1);
	update(0,0,n,1,0);
	for (int i=1;i<=n;i++) {
		int tmp=query(i-1,i-1,0,n,1)+s[i-1]+(b[i]==0);
		for (int j=0;j<g[i].size();j++)
			dp[j]=s[g[i][j]]+query(i-1,g[i][j],0,n,1);
		update(i,0,n,1,tmp-s[i]);
		for (int j=0;j<g[i].size();j++)
			update(g[i][j],0,n,1,dp[j]-s[g[i][j]]);
	}
	ans=query(n,n,0,n,1)+s[n];
	printf("%d\n",n-ans);
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn=200010;
const int INF=(1<<30);
int n,m,T[maxn*4],b[maxn],s[maxn],dp[maxn],ans;
vector<int> g[maxn];
void build(int l,int r,int root) {
	T[root]=-(1<<30);
	if (l==r) return;
	int mid=(l+r)/2;
	build(l,mid,root*2);
	build(mid+1,r,root*2+1);
}
int query(int L,int R,int l,int r,int root) {
	if (L<=l&&r<=R) return T[root];
	int mid=(l+r)/2,res=-INF;
	if (L<=mid) res=max(res,query(L,R,l,mid,root*2));
	if (mid<R) res=max(res,query(L,R,mid+1,r,root*2+1));
	return res;
}
void update(int pos,int l,int r,int root,int val) {
	if (l==r) {
		T[root]=max(T[root],val); return;
	}
	int mid=(l+r)/2;
	if (pos<=mid) update(pos,l,mid,root*2,val);
	else update(pos,mid+1,r,root*2+1,val);
	T[root]=max(T[root*2],T[root*2+1]);
}
int main() {
	//freopen("1.txt","r",stdin);
	scanf("%d",&n);
	for (int i=1;i<=n;i++) {
		scanf("%d",&b[i]);
		s[i]=s[i-1]+b[i];
	}
	scanf("%d",&m);
	for (int i=1;i<=m;i++) {
		int l,r; scanf("%d %d",&l,&r);
		g[l].push_back(r);
	}
	build(0,n,1);
	update(0,0,n,1,0);
	for (int i=1;i<=n;i++) {
		int tmp=query(i-1,i-1,0,n,1)+s[i-1]+(b[i]==0);
		for (int j=0;j<g[i].size();j++)
			dp[j]=s[g[i][j]]+query(i-1,g[i][j],0,n,1);
		update(i,0,n,1,tmp-s[i]);
		for (int j=0;j<g[i].size();j++)
			update(g[i][j],0,n,1,dp[j]-s[g[i][j]]);
	}
	ans=query(n,n,0,n,1)+s[n];
	printf("%d\n",n-ans);
	return 0;
}
```