//#pragma optimize("Ofast")
#include<bits/stdc++.h>
#define MAXN 200005
#define inf int(16843009)
using namespace std;
typedef long long ll;


int N,M;

int a[MAXN], b[MAXN];
int dp[MAXN];
vector<int> adj[MAXN];

struct Node{
	int l,r;
	int d;
	Node *ls, *rs;
	
	void update(){
		d = min(ls->d, rs->d);
	}
} pool[2*MAXN], *rt;

int top = 0;

Node* buildT(int l, int r){
	Node *p = pool + (++top);
	p->l = l; p->r = r;
	if(l==r){
		p->d = dp[l];
		return p;
	}
	
	int mid = (l+r)/2;
	p->ls = buildT(l, mid);
	p->rs = buildT(mid+1, r);
	
	p->update();
	return p;
}

void change(Node *p, int id, int x){
	if(p->l==id && p->r==id){
		p->d = min(p->d, x);
		return;
	}
	
	int mid = (p->l + p->r)/2;
	if(id <= mid) change(p->ls, id, x);
	else change(p->rs, id, x);
	
	p->update();
}

int query(Node *p, int l, int r){
	if(p->l==l && p->r==r){
		return p->d;
	}
	
	int mid = (p->l + p->r)/2;
	if(r <= mid) return query(p->ls, l, r);
	else if(l >= mid+1) return query(p->rs, l, r);
	else return min(query(p->ls, l, mid), query(p->rs, mid+1, r));
}


int main(){
	
	cin>>N;
	
	int sum0 = 0;
	for(int i=1;i<=N;i++){
		cin>>b[i];
		sum0 += (b[i]==0);
	}
	
	cin>>M;
	
	int l,r;
	for(int i=1;i<=M;i++){
		cin>>l>>r;
		adj[l].push_back(r);
	}
	
	memset(dp, 1, sizeof(dp));
	dp[0] = 0;
	rt = buildT(0,N);
	
	for(l=1;l<=N;l++){
		//
		int m;
		for(int k=0;k<adj[l].size();k++){
			r = adj[l][k];
			m = query(rt, l-1, r);
			if(m < dp[r]) change(rt, r, m);
			dp[r] = min(dp[r], m);
		}
		dp[l] = min(dp[l], dp[l-1] + (b[l]==1?1:-1));
		change(rt, l, dp[l]);
	}
	cerr<<"dp[N]: "<<dp[N]<<endl;
	cout<<sum0 + dp[N]<<endl;
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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
//#pragma optimize("Ofast")
#include<bits/stdc++.h>
#define MAXN 200005
#define inf int(16843009)
using namespace std;
typedef long long ll;


int N,M;

int a[MAXN], b[MAXN];
int dp[MAXN];
vector<int> adj[MAXN];

struct Node{
	int l,r;
	int d;
	Node *ls, *rs;
	
	void update(){
		d = min(ls->d, rs->d);
	}
} pool[2*MAXN], *rt;

int top = 0;

Node* buildT(int l, int r){
	Node *p = pool + (++top);
	p->l = l; p->r = r;
	if(l==r){
		p->d = dp[l];
		return p;
	}
	
	int mid = (l+r)/2;
	p->ls = buildT(l, mid);
	p->rs = buildT(mid+1, r);
	
	p->update();
	return p;
}

void change(Node *p, int id, int x){
	if(p->l==id && p->r==id){
		p->d = min(p->d, x);
		return;
	}
	
	int mid = (p->l + p->r)/2;
	if(id <= mid) change(p->ls, id, x);
	else change(p->rs, id, x);
	
	p->update();
}

int query(Node *p, int l, int r){
	if(p->l==l && p->r==r){
		return p->d;
	}
	
	int mid = (p->l + p->r)/2;
	if(r <= mid) return query(p->ls, l, r);
	else if(l >= mid+1) return query(p->rs, l, r);
	else return min(query(p->ls, l, mid), query(p->rs, mid+1, r));
}


int main(){
	
	cin>>N;
	
	int sum0 = 0;
	for(int i=1;i<=N;i++){
		cin>>b[i];
		sum0 += (b[i]==0);
	}
	
	cin>>M;
	
	int l,r;
	for(int i=1;i<=M;i++){
		cin>>l>>r;
		adj[l].push_back(r);
	}
	
	memset(dp, 1, sizeof(dp));
	dp[0] = 0;
	rt = buildT(0,N);
	
	for(l=1;l<=N;l++){
		//
		int m;
		for(int k=0;k<adj[l].size();k++){
			r = adj[l][k];
			m = query(rt, l-1, r);
			if(m < dp[r]) change(rt, r, m);
			dp[r] = min(dp[r], m);
		}
		dp[l] = min(dp[l], dp[l-1] + (b[l]==1?1:-1));
		change(rt, l, dp[l]);
	}
	cerr<<"dp[N]: "<<dp[N]<<endl;
	cout<<sum0 + dp[N]<<endl;
	return 0;
}
```