#include <bits/stdc++.h>
using namespace std;
const int N = 4e5 + 10, M = N << 2; //因为需要装无向图,所以N=2e5,又因为有新图有旧图,都保存到一个数组中,所以需要再乘以2
//链式前向星
int e[M], h1[N], h2[N], idx, w[M], ne[M];
void add(int h[], int a, int b, int c = 0) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
int f[N][16];
int depth[N], dist[N];
void bfs() {
// 1号点是源点
depth[1] = 1;
queue<int> q;
q.push(1);
while (q.size()) {
int u = q.front();
q.pop();
for (int i = h2[u]; ~i; i = ne[i]) {
int j = e[i];
if (!depth[j]) {
q.push(j);
depth[j] = depth[u] + 1;
dist[j] = dist[u] + w[i];
f[j][0] = u;
for (int k = 1; k <= 15; k++) f[j][k] = f[f[j][k - 1]][k - 1];
}
}
}
}
//最近公共祖先
int lca(int a, int b) {
if (depth[a] < depth[b]) swap(a, b);
for (int k = 15; k >= 0; k--)
if (depth[f[a][k]] >= depth[b]) a = f[a][k];
if (a == b) return a;
for (int k = 15; k >= 0; k--)
if (f[a][k] != f[b][k]) a = f[a][k], b = f[b][k];
return f[a][0];
}
//边双连通分量
int dfn[N], low[N], ts, stk[N], top;
vector<int> dcc[N]; //边双中有哪些原始点
int id[N], dcc_cnt; //原始点x属于哪个边双连通分量,dcc_cnt指边双连通分量个数
int is_bridge[M]; //记录哪些边是割边
void tarjan(int u, int fa) {
dfn[u] = low[u] = ++ts;
stk[++top] = u;
for (int i = h1[u]; ~i; i = ne[i]) {
int j = e[i];
if (!dfn[j]) {
tarjan(j, i);
low[u] = min(low[u], low[j]);
if (dfn[u] < low[j]) is_bridge[i] = is_bridge[i ^ 1] = true; //记录割边
} else if (i != (fa ^ 1))
low[u] = min(low[u], dfn[j]);
}
if (dfn[u] == low[u]) {
++dcc_cnt; //边双数量+1
int x;
do {
x = stk[top--];
id[x] = dcc_cnt; // 记录点与边双关系
dcc[dcc_cnt].push_back(x); // 记录边双中有哪些点
} while (x != u);
}
}
int n, m, q;
int main() {
memset(h1, -1, sizeof h1); // h1是原图的表头
memset(h2, -1, sizeof h2); // h2是新生成的缩完点的图表头
scanf("%d %d", &n, &m);
//原图
for (int i = 1; i <= m; i++) {
int a, b;
scanf("%d %d", &a, &b);
add(h1, a, b), add(h1, b, a);
}
//用tarjan来缩点,将边双连通分量缩点
for (int i = 1; i <= n; i++)
if (!dfn[i]) tarjan(i, -1);
//将新图建出来
for (int u = 1; u <= n; u++)
for (int i = h1[u]; ~i; i = ne[i]) {
int j = e[i];
if (id[u] != id[j])
add(h2, id[u], id[j]), add(h2, id[j], id[u]);
}
//随便找个存在的号作为根节点,预处理出每个点到根节点的距离
bfs();
scanf("%d", &q); // q次询问
for (int i = 1; i <= q; i++) {
int a, b;
scanf("%d %d", &a, &b);
a = id[a], b = id[b];
printf("%d\n", depth[a] + depth[b] - depth[lca(a, b)] * 2); //这就很显然了
}
return 0;
}