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

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

int read()
{
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)) { if(c == '-') f = 0; c = getchar(); }
	while(isdigit(c)) { x = (x << 3) + (x << 1) + (c ^ 48); c = getchar(); }
	return f ? x : -x;
}

const int maxn = 2e5 + 10;

struct SGT
{
	int t[maxn << 2];

	void upt(int o, int l, int r, int x, int v)
	{
		if(l == r) { t[o] = min(t[o], v); return ; } 
		int mid = (l + r) >> 1;
		if(x <= mid) upt(o << 1, l, mid, x, v);
		else upt(o << 1 | 1, mid + 1, r, x, v);
		t[o] = min(t[o << 1 | 1], t[o << 1]);
	}

	int que(int o, int l, int r, int x, int y)
	{
		if(x <= l && r <= y) return t[o]; 
		int mid = (l + r) >> 1, ret = 0x3f3f3f3f;
		if(x <= mid) ret = que(o << 1, l, mid, x, y);
		if(y > mid) ret = min(ret, que(o << 1 | 1, mid + 1, r, x, y));
		return ret;
	}
}S1, S2;

int b[maxn], sum0[maxn], sum1[maxn];
pair<int, int> Q[maxn];

int main()
{
	memset(S1.t, 0x3f, sizeof(S1.t));
	memset(S2.t, 0x3f, sizeof(S2.t));

	int n = read();
	for(int i = 1; i <= n; ++i) b[i] = read();
	for(int i = 1; i <= n; ++i) sum0[i] = sum0[i - 1] + (b[i] == 0), sum1[i] = sum1[i - 1] + (b[i] == 1);

	int m = read();
	for(int i = 1; i <= m; ++i)
	{
		int x = read(), y = read();
		Q[i] = make_pair(x, y);
	}
	sort(Q + 1, Q + m + 1);

	sum0[n + 1] = sum0[n];
	Q[m + 1] = make_pair(n + 1, n + 1); ++m;
	S1.upt(1, 0, n, 0, 0);
	S2.upt(1, 0, n, 0, 0);
	for(int i = 1; i <= m; ++i)
	{
		int tmp = S2.que(1, 0, n, 0, Q[i].first - 1);
		//cout << '*' << tmp << ' ' << Q[i].first - 1 << endl;
		tmp = tmp + sum1[Q[i].first - 1] + sum0[Q[i].second] - sum0[Q[i].first - 1];
		if(i == m) { printf("%d\n", tmp); return 0; }
		int tmp1 = S1.que(1, 0, n, Q[i].first, Q[i].second);
		tmp1 = tmp1 + sum0[Q[i].second];
		int now = min(tmp, tmp1);
		//cout << Q[i].second << ' ' << now - sum1[Q[i].second] << endl;
		S1.upt(1, 0, n, Q[i].second, now - sum0[Q[i].second]);
		S2.upt(1, 0, n, Q[i].second, now - sum1[Q[i].second]);
	}

	return 0;
}