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 <bits/stdc++.h>
using namespace std;
const int maxn = 202000;
int pref[maxn];
vector<int> e[maxn];
int sum(int l, int r){
	return pref[r] - pref[l - 1];
}
int mx[maxn<<2], lazy[maxn<<2];
void push(int id, int l, int r){
	if(lazy[id]){
		mx[id] += lazy[id];
		if(l != r){
			lazy[id<<1] += lazy[id];
			lazy[id<<1|1] += lazy[id];
		}
		lazy[id] = 0;
	}
}
void add(int id, int l, int r, int x, int y, int v){
	push(id, l, r);
	if(l > y || r < x) return;
	if(x <= l && r <= y){
		lazy[id] += v;
		push(id, l, r);
		return;
	}
	int mid = (l + r)>>1;
	add(id<<1, l, mid, x, y, v);
	add(id<<1|1, mid + 1, r, x, y, v);
	mx[id] = max(mx[id<<1], mx[id<<1|1]);
}
void add_all(int l, int r, int x){
	add(1, 0, maxn, l, r, x);
}
void upd(int id, int l, int r, int p, int v){
	push(id, l, r);
	if(l > p || r < p) return;
	if(p <= l && r <= p){
		mx[id] = max(mx[id], v);
		return;
	}
	int mid = (l+r)>>1;
	upd(id<<1, l, mid, p, v);
	upd(id<<1|1, mid + 1, r, p, v);
	mx[id] = max(mx[id<<1], mx[id<<1|1]);
}
void upd_max(int p, int x){
	upd(1, 0, maxn, p, x);
}
int get(int id, int l, int r, int x, int y){
	push(id, l, r);
	if(l > y || r < x) return -(1<<30);
	if(x <= l && r <= y) return mx[id];
	int mid = (l + r)>>1;
	return max(get(id<<1, l, mid, x, y), get(id<<1|1, mid + 1, r, x, y));
}
int get_max(int l, int r){
	return get(1, 0, maxn, l, r);
}
int main(){
	int n; cin >> n;
	for(int i = 0, x; i < n; i++){
		cin >> x;
		pref[i + 1] = pref[i] + x;
	}
	int q; cin >> q;
	for(int i = 0; i < q; i++){
		int l, r; cin >> l >> r;
		e[l].push_back(r);
	}
	for(int i = 1; i <= n; i++){
		add_all(i, i, -pref[i]);
	}
	for(int i = 1; i <= n; i++){
		for(int to : e[i]){
			int best = get_max(i, to);
			upd_max(to, best);
			int opt2 = get_max(0, i - 1) - pref[i - 1];
			upd_max(to, opt2);
		}
		if(sum(i, i) == 0){
			add_all(0, i - 1, 1);
		}
		add_all(i, i, pref[i]);
	}
	cout << n - get_max(0, n) << endl;
	return 0;
}