#include<bits/stdc++.h>
using namespace std;

const int maxn=200010;
const int inf=2e9;
int n,q;
int b[maxn];
int ans;
int sum[maxn];
vector<int> g[maxn];
int dp[maxn];

void read(int& x){
	x=0; char c=getchar();
	while(c<'0' || c>'9') c=getchar();
	while(c>='0' && c<='9') x=x*10+c-'0', c=getchar();
}

void chkmin(int& x,int y){ if(x>y) x=y; }

int Min[maxn<<2];
void pushup(int o){
	Min[o]=min(Min[o<<1],Min[o<<1|1]);
}
void build(int o,int l,int r){
	if(l==r){ Min[o]=dp[l]-sum[l]; return; }
	int mid=(l+r)>>1;
	build(o<<1,l,mid); build(o<<1|1,mid+1,r);
	pushup(o);
}
int query(int o,int l,int r,int ql,int qr){
	if(ql<=l && qr>=r){ return Min[o]; }
	int mid=(l+r)>>1, ret=inf;
	if(ql<=mid) ret=query(o<<1,l,mid,ql,qr); if(qr>mid) chkmin(ret, query(o<<1|1,mid+1,r,ql,qr));
	return ret;
}
void modify(int o,int l,int r,int pos){
	if(l==r){ Min[o]=dp[l]-sum[l]; return; }
	int mid=(l+r)>>1;
	if(pos<=mid) modify(o<<1,l,mid,pos); else modify(o<<1|1,mid+1,r,pos);
	pushup(o);
}

int main(){

	read(n);
	for(int i=1;i<=n;i++) read(b[i]), sum[i]=sum[i-1]+(b[i]^1);
	read(q);
	int l, r;
	for(int i=1;i<=q;i++) read(l), read(r), g[l-1].push_back(r);
	for(int i=1;i<=n;i++) dp[i]=inf;
	build(1,0,n);
	for(int i=0;i<=n;i++){
		if(i>0) chkmin(dp[i], dp[i-1]+(b[i])), modify(1,0,n,i);
		for(int j=0;j<g[i].size();j++){
			chkmin(dp[g[i][j]], query(1,0,n,i,g[i][j]-1)+sum[g[i][j]]);
			modify(1,0,n,g[i][j]);
		}
		// printf("%d : %d\n", i, dp[i]);
	}
	printf("%d\n", dp[n]);
	return 0;
}
### Prompt

Your task is to create a Cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

const int maxn=200010;
const int inf=2e9;
int n,q;
int b[maxn];
int ans;
int sum[maxn];
vector<int> g[maxn];
int dp[maxn];

void read(int& x){
	x=0; char c=getchar();
	while(c<'0' || c>'9') c=getchar();
	while(c>='0' && c<='9') x=x*10+c-'0', c=getchar();
}

void chkmin(int& x,int y){ if(x>y) x=y; }

int Min[maxn<<2];
void pushup(int o){
	Min[o]=min(Min[o<<1],Min[o<<1|1]);
}
void build(int o,int l,int r){
	if(l==r){ Min[o]=dp[l]-sum[l]; return; }
	int mid=(l+r)>>1;
	build(o<<1,l,mid); build(o<<1|1,mid+1,r);
	pushup(o);
}
int query(int o,int l,int r,int ql,int qr){
	if(ql<=l && qr>=r){ return Min[o]; }
	int mid=(l+r)>>1, ret=inf;
	if(ql<=mid) ret=query(o<<1,l,mid,ql,qr); if(qr>mid) chkmin(ret, query(o<<1|1,mid+1,r,ql,qr));
	return ret;
}
void modify(int o,int l,int r,int pos){
	if(l==r){ Min[o]=dp[l]-sum[l]; return; }
	int mid=(l+r)>>1;
	if(pos<=mid) modify(o<<1,l,mid,pos); else modify(o<<1|1,mid+1,r,pos);
	pushup(o);
}

int main(){

	read(n);
	for(int i=1;i<=n;i++) read(b[i]), sum[i]=sum[i-1]+(b[i]^1);
	read(q);
	int l, r;
	for(int i=1;i<=q;i++) read(l), read(r), g[l-1].push_back(r);
	for(int i=1;i<=n;i++) dp[i]=inf;
	build(1,0,n);
	for(int i=0;i<=n;i++){
		if(i>0) chkmin(dp[i], dp[i-1]+(b[i])), modify(1,0,n,i);
		for(int j=0;j<g[i].size();j++){
			chkmin(dp[g[i][j]], query(1,0,n,i,g[i][j]-1)+sum[g[i][j]]);
			modify(1,0,n,g[i][j]);
		}
		// printf("%d : %d\n", i, dp[i]);
	}
	printf("%d\n", dp[n]);
	return 0;
}
```