#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7, siz = 1e6 + 5;
vector<long long> g[siz];
long long n, t, m, k, a[siz], sum[siz], l[siz], vis[siz];
stack<long long> st;
void go(long long u, long long p) {
  vis[u] = 1;
  st.push(u);
  for (auto v : g[u]) {
    if (v == p) continue;
    if (!vis[v])
      go(v, u);
    else if (vis[v] == 1) {
      stack<long long> cur;
      long long mn = v, mx = v;
      while (st.top() != v) {
        mn = min(mn, st.top());
        mx = max(mx, st.top());
        cur.push(st.top());
        st.pop();
      }
      l[mx] = mn;
      while (cur.size()) st.push(cur.top()), cur.pop();
    }
  }
  if (st.size()) st.pop();
  vis[u] = 2;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 1; i <= m; i++) {
    long long x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (long long i = 1; i <= n; i++)
    if (!vis[i]) go(i, 0);
  for (long long i = 1; i <= n; i++) {
    l[i] = max(l[i - 1], l[i]);
    sum[i] = sum[i - 1] + i - l[i];
  }
  long long q;
  cin >> q;
  while (q--) {
    long long lft, rit;
    cin >> lft >> rit;
    long long pos_left = lower_bound(l + lft, l + 1 + rit, lft) - l;
    long long ans = (pos_left - lft) * (pos_left - lft + 1) / 2;
    ans += sum[rit] - sum[pos_left - 1];
    cout << ans << endl;
  }
}

### Prompt

Generate a Cpp solution to the following problem:

You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7, siz = 1e6 + 5;
vector<long long> g[siz];
long long n, t, m, k, a[siz], sum[siz], l[siz], vis[siz];
stack<long long> st;
void go(long long u, long long p) {
  vis[u] = 1;
  st.push(u);
  for (auto v : g[u]) {
    if (v == p) continue;
    if (!vis[v])
      go(v, u);
    else if (vis[v] == 1) {
      stack<long long> cur;
      long long mn = v, mx = v;
      while (st.top() != v) {
        mn = min(mn, st.top());
        mx = max(mx, st.top());
        cur.push(st.top());
        st.pop();
      }
      l[mx] = mn;
      while (cur.size()) st.push(cur.top()), cur.pop();
    }
  }
  if (st.size()) st.pop();
  vis[u] = 2;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 1; i <= m; i++) {
    long long x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (long long i = 1; i <= n; i++)
    if (!vis[i]) go(i, 0);
  for (long long i = 1; i <= n; i++) {
    l[i] = max(l[i - 1], l[i]);
    sum[i] = sum[i - 1] + i - l[i];
  }
  long long q;
  cin >> q;
  while (q--) {
    long long lft, rit;
    cin >> lft >> rit;
    long long pos_left = lower_bound(l + lft, l + 1 + rit, lft) - l;
    long long ans = (pos_left - lft) * (pos_left - lft + 1) / 2;
    ans += sum[rit] - sum[pos_left - 1];
    cout << ans << endl;
  }
}

```