#include <bits/stdc++.h>

using namespace std;

const int maxn = 2e5 + 7;

struct Que{
	int l,r;
	bool operator < (const Que& a)const{
		return (l == a.l) ? (r < a.r) : (l < a.l);
	}
}q[maxn];
int Min[maxn << 2];
int tag[maxn << 2];
int a[maxn];
int sum;
int n,Q;
int Read(char c = 0,int re = 0){
	while(!isdigit(c = getchar()));
	re = c - '0';
	while(isdigit(c = getchar()))
		re = ((re << 2) + re << 1) + c - '0';
	return re;
}
#define MID() int mid = l + r >> 1
void Pushdown(int u){
	int add = tag[u];tag[u] = 0;
	Min[u << 1] += add;
	Min[u << 1 | 1] += add;
	tag[u << 1] += add;
	tag[u << 1 | 1] += add;
}
void Update(int u,int l,int r,int L,int R,int v){
	if(l >= L && r <= R){
		Min[u] += v;
		tag[u] += v;
		return;
	}
	MID();
	if(tag[u])Pushdown(u);
	if(L <= mid)Update(u << 1,l,mid,L,R,v);
	if(R >  mid)Update(u << 1 | 1,mid + 1,r,L,R,v);
	Min[u] = min(Min[u << 1],Min[u << 1 | 1]);
}
void Modify(int u,int l,int r,int x,int v){
	if(l == r)return Min[u] = min(Min[u],v),void();
	MID();
	if(x <= mid)Modify(u << 1,l,mid,x,v);
	else Modify(u << 1 | 1,mid + 1,r,x,v);
	Min[u] = min(Min[u << 1],Min[u << 1 | 1]);
}
int Query(int u,int l,int r,int L,int R){
	if(l >= L && r <= R)return Min[u];
	if(tag[u])Pushdown(u);
	MID();
	int re = 0x3f3f3f3f;
	if(L <= mid)re = min(re,Query(u << 1,l,mid,L,R));
	if(R >  mid)re = min(re,Query(u << 1 | 1,mid + 1,r,L,R));
	return re;
}
int main(){
	n = Read();
	memset(Min,0x3f,sizeof Min);
	Modify(1,0,n,0,0);
	for(int i = 1; i <= n; ++i)
		a[i] = Read(),sum += (a[i] == 0);
	Q = Read();
	for(int i = 1; i <= Q; ++i)
		q[i].l = Read(),q[i].r = Read();
	sort(q + 1,q + Q + 1);
	int cur = 1;
	for(int i = 1; i <= n; ++i){
		while(q[cur].l == i){
			int tmp = Query(1,0,n,0,q[cur].r - 1);
			Modify(1,0,n,q[cur].r,tmp);
			cur++;
		}
		Update(1,0,n,0,i - 1,(a[i] == 1) ? 1 : -1);
	}
	printf("%d\n",Query(1,0,n,0,n) + sum);
	return 0;
}

### Prompt

Please formulate 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 = 2e5 + 7;

struct Que{
	int l,r;
	bool operator < (const Que& a)const{
		return (l == a.l) ? (r < a.r) : (l < a.l);
	}
}q[maxn];
int Min[maxn << 2];
int tag[maxn << 2];
int a[maxn];
int sum;
int n,Q;
int Read(char c = 0,int re = 0){
	while(!isdigit(c = getchar()));
	re = c - '0';
	while(isdigit(c = getchar()))
		re = ((re << 2) + re << 1) + c - '0';
	return re;
}
#define MID() int mid = l + r >> 1
void Pushdown(int u){
	int add = tag[u];tag[u] = 0;
	Min[u << 1] += add;
	Min[u << 1 | 1] += add;
	tag[u << 1] += add;
	tag[u << 1 | 1] += add;
}
void Update(int u,int l,int r,int L,int R,int v){
	if(l >= L && r <= R){
		Min[u] += v;
		tag[u] += v;
		return;
	}
	MID();
	if(tag[u])Pushdown(u);
	if(L <= mid)Update(u << 1,l,mid,L,R,v);
	if(R >  mid)Update(u << 1 | 1,mid + 1,r,L,R,v);
	Min[u] = min(Min[u << 1],Min[u << 1 | 1]);
}
void Modify(int u,int l,int r,int x,int v){
	if(l == r)return Min[u] = min(Min[u],v),void();
	MID();
	if(x <= mid)Modify(u << 1,l,mid,x,v);
	else Modify(u << 1 | 1,mid + 1,r,x,v);
	Min[u] = min(Min[u << 1],Min[u << 1 | 1]);
}
int Query(int u,int l,int r,int L,int R){
	if(l >= L && r <= R)return Min[u];
	if(tag[u])Pushdown(u);
	MID();
	int re = 0x3f3f3f3f;
	if(L <= mid)re = min(re,Query(u << 1,l,mid,L,R));
	if(R >  mid)re = min(re,Query(u << 1 | 1,mid + 1,r,L,R));
	return re;
}
int main(){
	n = Read();
	memset(Min,0x3f,sizeof Min);
	Modify(1,0,n,0,0);
	for(int i = 1; i <= n; ++i)
		a[i] = Read(),sum += (a[i] == 0);
	Q = Read();
	for(int i = 1; i <= Q; ++i)
		q[i].l = Read(),q[i].r = Read();
	sort(q + 1,q + Q + 1);
	int cur = 1;
	for(int i = 1; i <= n; ++i){
		while(q[cur].l == i){
			int tmp = Query(1,0,n,0,q[cur].r - 1);
			Modify(1,0,n,q[cur].r,tmp);
			cur++;
		}
		Update(1,0,n,0,i - 1,(a[i] == 1) ? 1 : -1);
	}
	printf("%d\n",Query(1,0,n,0,n) + sum);
	return 0;
}

```