#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
const int maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
using namespace std;
struct T{
	struct A{
		int l, r, Min;
	}a[maxn << 2];
	void build(int cur, int l, int r){
		a[cur].l = l, a[cur].r = r;
		a[cur].Min = INF;
		if (l == r) return;
		int mid = l + r >> 1;
		build(cur << 1, l, mid);
		build(cur << 1 | 1, mid + 1, r);
	}
	void upd(int cur, int p, int k){
		if (a[cur].l == a[cur].r){
			a[cur].Min = min(a[cur].Min, k);
			return;
		}
		int mid = a[cur].l + a[cur].r >> 1;
		if (p <= mid) upd(cur << 1, p, k);
		else upd(cur << 1 | 1, p, k);
		a[cur].Min = min(a[cur << 1].Min, a[cur << 1 | 1].Min);
	}
	int Query(int cur, int l, int r){
		if (a[cur].l > r || a[cur].r < l) return INF;
		if (a[cur].l >= l && a[cur].r <= r) return a[cur].Min;
		return min(Query(cur << 1, l, r), Query(cur << 1 | 1, l, r));
	}
}t;
vector <int> v[maxn];
int n, Q, b[maxn], f[maxn], S = 0;
int main(){
	scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%d", b + i), S += (1 - b[i]);
	scanf("%d", &Q);
	for (int l, r, i = 1; i <= Q; i++){
		scanf("%d%d", &l, &r);
		v[l].push_back(r);
	} t.build(1, 1, n);
	for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
	memset(f, 0x3f, sizeof f); f[0] = 0;
	for (int i = 1; i <= n; i++){
		for (int j = 0; j < v[i].size(); j++){
			int r = v[i][j]; int tmp = min(f[i - 1], t.Query(1, max(1, i - 1), r));
			if (tmp < f[r]) t.upd(1, r, f[r] = tmp);
		}
		f[i] = min(f[i], f[i - 1] + ((b[i] == 1) ? 1 : -1));
	} printf("%d\n", S + f[n]);
	return 0;
}
### Prompt

In CPP, your task is to solve 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 <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
const int maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
using namespace std;
struct T{
	struct A{
		int l, r, Min;
	}a[maxn << 2];
	void build(int cur, int l, int r){
		a[cur].l = l, a[cur].r = r;
		a[cur].Min = INF;
		if (l == r) return;
		int mid = l + r >> 1;
		build(cur << 1, l, mid);
		build(cur << 1 | 1, mid + 1, r);
	}
	void upd(int cur, int p, int k){
		if (a[cur].l == a[cur].r){
			a[cur].Min = min(a[cur].Min, k);
			return;
		}
		int mid = a[cur].l + a[cur].r >> 1;
		if (p <= mid) upd(cur << 1, p, k);
		else upd(cur << 1 | 1, p, k);
		a[cur].Min = min(a[cur << 1].Min, a[cur << 1 | 1].Min);
	}
	int Query(int cur, int l, int r){
		if (a[cur].l > r || a[cur].r < l) return INF;
		if (a[cur].l >= l && a[cur].r <= r) return a[cur].Min;
		return min(Query(cur << 1, l, r), Query(cur << 1 | 1, l, r));
	}
}t;
vector <int> v[maxn];
int n, Q, b[maxn], f[maxn], S = 0;
int main(){
	scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%d", b + i), S += (1 - b[i]);
	scanf("%d", &Q);
	for (int l, r, i = 1; i <= Q; i++){
		scanf("%d%d", &l, &r);
		v[l].push_back(r);
	} t.build(1, 1, n);
	for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
	memset(f, 0x3f, sizeof f); f[0] = 0;
	for (int i = 1; i <= n; i++){
		for (int j = 0; j < v[i].size(); j++){
			int r = v[i][j]; int tmp = min(f[i - 1], t.Query(1, max(1, i - 1), r));
			if (tmp < f[r]) t.upd(1, r, f[r] = tmp);
		}
		f[i] = min(f[i], f[i - 1] + ((b[i] == 1) ? 1 : -1));
	} printf("%d\n", S + f[n]);
	return 0;
}
```