#include <bits/stdc++.h>
using namespace std;
int now[5 * 100010], after[5 * 100010], mx[5 * 100010];
int lft[5 * 100010], rgt[5 * 100010], vi[5 * 100010], timer = 0, comp = 0;
long long Add[5 * 100010];
vector<int> g[5 * 100010];
stack<int> stk;
inline void dfs(int c, int p) {
  vi[c] = 1;
  now[c] = after[c] = ++timer;
  stk.push(c);
  for (auto x : g[c]) {
    if (x != p) {
      if (!vi[x]) {
        dfs(x, c);
        now[c] = min(now[c], now[x]);
      } else
        now[c] = min(now[c], now[x]);
    }
  }
  if (now[c] == after[c]) {
    comp++;
    lft[comp] = rgt[comp] = c;
    while (1) {
      int node = stk.top();
      stk.pop();
      lft[comp] = min(lft[comp], node);
      rgt[comp] = max(rgt[comp], node);
      if (node == c) break;
    }
  }
}
int main() {
  int n, m;
  scanf("%d", &n);
  ;
  scanf("%d", &m);
  ;
  memset(vi, 0, sizeof(vi));
  ;
  for (int x = 1; x <= m; x++) {
    int u, v;
    scanf("%d", &u);
    ;
    scanf("%d", &v);
    ;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  for (int x = 1; x <= n; x++)
    if (!vi[x]) dfs(x, 0);
  for (int x = 1; x <= n; x++) mx[x] = n;
  for (int x = 1; x <= comp; x++)
    if (lft[x] != rgt[x]) mx[lft[x]] = rgt[x] - 1;
  for (int x = n - 1; x >= 1; x--) mx[x] = min(mx[x + 1], mx[x]);
  for (int x = 1; x <= n; x++) Add[x] = Add[x - 1] + mx[x] - x + 1;
  int q;
  scanf("%d", &q);
  ;
  while (q--) {
    int from, to;
    scanf("%d", &from);
    ;
    scanf("%d", &to);
    ;
    int l = from - 1, r = to + 1, chk = from;
    while (l + 1 < r) {
      int m = l + r >> 1;
      if (mx[m] > to)
        r = m;
      else
        chk = max(chk, m), l = m;
    }
    chk = l;
    printf("%lld\n",
           Add[chk] - Add[from - 1] + 1LL * (to - chk) * (to - chk + 1LL) / 2);
  }
}

### Prompt

Please formulate 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;
int now[5 * 100010], after[5 * 100010], mx[5 * 100010];
int lft[5 * 100010], rgt[5 * 100010], vi[5 * 100010], timer = 0, comp = 0;
long long Add[5 * 100010];
vector<int> g[5 * 100010];
stack<int> stk;
inline void dfs(int c, int p) {
  vi[c] = 1;
  now[c] = after[c] = ++timer;
  stk.push(c);
  for (auto x : g[c]) {
    if (x != p) {
      if (!vi[x]) {
        dfs(x, c);
        now[c] = min(now[c], now[x]);
      } else
        now[c] = min(now[c], now[x]);
    }
  }
  if (now[c] == after[c]) {
    comp++;
    lft[comp] = rgt[comp] = c;
    while (1) {
      int node = stk.top();
      stk.pop();
      lft[comp] = min(lft[comp], node);
      rgt[comp] = max(rgt[comp], node);
      if (node == c) break;
    }
  }
}
int main() {
  int n, m;
  scanf("%d", &n);
  ;
  scanf("%d", &m);
  ;
  memset(vi, 0, sizeof(vi));
  ;
  for (int x = 1; x <= m; x++) {
    int u, v;
    scanf("%d", &u);
    ;
    scanf("%d", &v);
    ;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  for (int x = 1; x <= n; x++)
    if (!vi[x]) dfs(x, 0);
  for (int x = 1; x <= n; x++) mx[x] = n;
  for (int x = 1; x <= comp; x++)
    if (lft[x] != rgt[x]) mx[lft[x]] = rgt[x] - 1;
  for (int x = n - 1; x >= 1; x--) mx[x] = min(mx[x + 1], mx[x]);
  for (int x = 1; x <= n; x++) Add[x] = Add[x - 1] + mx[x] - x + 1;
  int q;
  scanf("%d", &q);
  ;
  while (q--) {
    int from, to;
    scanf("%d", &from);
    ;
    scanf("%d", &to);
    ;
    int l = from - 1, r = to + 1, chk = from;
    while (l + 1 < r) {
      int m = l + r >> 1;
      if (mx[m] > to)
        r = m;
      else
        chk = max(chk, m), l = m;
    }
    chk = l;
    printf("%lld\n",
           Add[chk] - Add[from - 1] + 1LL * (to - chk) * (to - chk + 1LL) / 2);
  }
}

```