#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#define MAXN 200010
int n, a[MAXN], Q, pre0[MAXN], pre1[MAXN], tree[MAXN << 2], tree2[MAXN << 2];
void getmin(int & a, int b) {if (a > b) a = b;}
struct Query {
	int l, r;
	inline bool operator < (const Query & b) const {return l == b.l ? r < b.r : l < b.l;}
	void read() {scanf("%d%d", &l, &r);}
} qs[MAXN];
void modify(int u, int l, int r, int tar, int v) {
	if (l == r) {
		tree[u] = v;
		tree2[u] = v - pre0[tar] + pre1[tar];
		return ;
	}
	int mid = l + r >> 1;
	if (tar <= mid) modify(u << 1, l, mid, tar, v);
	else modify(u << 1 | 1, mid + 1, r, tar, v);
	tree[u] = min(tree[u << 1], tree[u << 1 | 1]);
	tree2[u] = min(tree2[u << 1], tree2[u << 1 | 1]); 
}
int query(int u, int l, int r, int L, int R, int * tree) {
	if (L <= l && r <= R) return tree[u];
	int mid = l + r >> 1, res = 0x3f3f3f3f;
	if (L <= mid) getmin(res, query(u << 1, l, mid, L, R, tree));
	if (mid < R) getmin(res, query(u << 1 | 1, mid + 1, r, L, R, tree));
	return res;
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
		pre0[i] = pre0[i - 1] + !a[i];
		pre1[i] = pre1[i - 1] + a[i];
	}
	scanf("%d", &Q);
	for (int i = 1; i <= Q; ++i) qs[i].read();
	sort(qs + 1, qs + 1 + Q);
	memset(tree, 0x3f, sizeof tree);
	memset(tree2, 0x3f, sizeof tree2);
	modify(1, 0, n, 0, pre1[n]);
	for (int i = 1; i <= Q; ++i) {
		int tx = min(	query(1, 0, n, 0, qs[i].l - 1, tree) - pre0[qs[i].l - 1] + pre1[qs[i].l - 1], 
						query(1, 0, n, qs[i].l, qs[i].r - 1, tree2))
				+ pre0[qs[i].r] - pre1[qs[i].r];
		getmin(tx, query(1, 0, n, qs[i].r, qs[i].r, tree));
		modify(1, 0, n, qs[i].r, tx);
	}
	printf("%d\n", query(1, 0, n, 0, n, tree));
	return 0;
}

### Prompt

Your challenge is to write 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 <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#define MAXN 200010
int n, a[MAXN], Q, pre0[MAXN], pre1[MAXN], tree[MAXN << 2], tree2[MAXN << 2];
void getmin(int & a, int b) {if (a > b) a = b;}
struct Query {
	int l, r;
	inline bool operator < (const Query & b) const {return l == b.l ? r < b.r : l < b.l;}
	void read() {scanf("%d%d", &l, &r);}
} qs[MAXN];
void modify(int u, int l, int r, int tar, int v) {
	if (l == r) {
		tree[u] = v;
		tree2[u] = v - pre0[tar] + pre1[tar];
		return ;
	}
	int mid = l + r >> 1;
	if (tar <= mid) modify(u << 1, l, mid, tar, v);
	else modify(u << 1 | 1, mid + 1, r, tar, v);
	tree[u] = min(tree[u << 1], tree[u << 1 | 1]);
	tree2[u] = min(tree2[u << 1], tree2[u << 1 | 1]); 
}
int query(int u, int l, int r, int L, int R, int * tree) {
	if (L <= l && r <= R) return tree[u];
	int mid = l + r >> 1, res = 0x3f3f3f3f;
	if (L <= mid) getmin(res, query(u << 1, l, mid, L, R, tree));
	if (mid < R) getmin(res, query(u << 1 | 1, mid + 1, r, L, R, tree));
	return res;
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
		pre0[i] = pre0[i - 1] + !a[i];
		pre1[i] = pre1[i - 1] + a[i];
	}
	scanf("%d", &Q);
	for (int i = 1; i <= Q; ++i) qs[i].read();
	sort(qs + 1, qs + 1 + Q);
	memset(tree, 0x3f, sizeof tree);
	memset(tree2, 0x3f, sizeof tree2);
	modify(1, 0, n, 0, pre1[n]);
	for (int i = 1; i <= Q; ++i) {
		int tx = min(	query(1, 0, n, 0, qs[i].l - 1, tree) - pre0[qs[i].l - 1] + pre1[qs[i].l - 1], 
						query(1, 0, n, qs[i].l, qs[i].r - 1, tree2))
				+ pre0[qs[i].r] - pre1[qs[i].r];
		getmin(tx, query(1, 0, n, qs[i].r, qs[i].r, tree));
		modify(1, 0, n, qs[i].r, tx);
	}
	printf("%d\n", query(1, 0, n, 0, n, tree));
	return 0;
}

```