Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const long long INF = (long long)5e18;
const int MOD = 998244353;
int _abs(int x) { return x < 0 ? -x : x; }
int add(int x, int y) {
  x += y;
  return x >= MOD ? x - MOD : x;
}
int sub(int x, int y) {
  x -= y;
  return x < 0 ? x + MOD : x;
}
void Add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
void Sub(int &x, int y) {
  x -= y;
  if (x < 0) x += MOD;
}
void Mul(int &x, int y) { x = (long long)(x) * (y) % MOD; }
int qpow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = (long long)(ret) * (x) % MOD;
    x = (long long)(x) * (x) % MOD;
    y >>= 1;
  }
  return ret;
}
void checkmin(int &x, int y) {
  if (x > y) x = y;
}
void checkmax(int &x, int y) {
  if (x < y) x = y;
}
void checkmin(long long &x, long long y) {
  if (x > y) x = y;
}
void checkmax(long long &x, long long y) {
  if (x < y) x = y;
}
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return x * f;
}
const int N = 4105;
const int M = 66000;
const int B = 256;
int n, a[N], pos[N], q, cnt, bel[N];
vector<pair<int, int> > myy;
int Merge(int x, int y) {
  myy.push_back(make_pair(x, y));
  cnt++;
  return cnt;
}
struct node {
  int l, r;
} Q[M];
struct node2 {
  int l, r, to;
};
vector<int> vl[N], vr[N];
int mp[N][N];
int b[N], top = 0;
void solve(int l, int r, vector<int> v) {
  if (v.empty()) return;
  if (l == r) {
    mp[v[0]][v[0]] = b[v[0]];
    return;
  }
  int mid = (l + r) >> 1;
  vector<int> pl, pr;
  for (auto &u : v) {
    if (a[b[u]] <= mid)
      pl.push_back(u);
    else
      pr.push_back(u);
  }
  solve(l, mid, pl);
  solve(mid + 1, r, pr);
  vector<node2> tt;
  for (int i = 0, j = 0; i < (int)(pl).size(); i++) {
    while (j < (int)(pr).size() && pr[j] < pl[i]) j++;
    int x = i + 1, y = j;
    while (x < (int)(pl).size() || y < (int)(pr).size()) {
      if (y == (int)(pr).size() || (x != (int)(pl).size() && pl[x] < pr[y]))
        x++;
      else
        y++;
      if (y != j)
        tt.push_back({pl[i], max(pl[x - 1], pr[y - 1]),
                      Merge(mp[pl[i]][pl[x - 1]], mp[pr[j]][pr[y - 1]])});
    }
  }
  for (int i = 0, j = 0; j < (int)(pr).size(); j++) {
    while (i < (int)(pl).size() && pl[i] < pr[j]) i++;
    int x = i, y = j + 1;
    while (x < (int)(pl).size() || y < (int)(pr).size()) {
      if (y == (int)(pr).size() || (x != (int)(pl).size() && pl[x] < pr[y]))
        x++;
      else
        y++;
      if (x != i)
        tt.push_back({pr[j], max(pl[x - 1], pr[y - 1]),
                      Merge(mp[pl[i]][pl[x - 1]], mp[pr[j]][pr[y - 1]])});
    }
  }
  for (auto &u : tt) mp[u.l][u.r] = u.to;
}
vector<int> v[M];
int L[M], R[M];
void run(int id) {
  top = 0;
  for (int i = 1; i <= n; i++)
    if (bel[i] == id) b[++top] = i;
  vector<int> tt;
  for (int i = 1; i <= top; i++) tt.push_back(i);
  solve((id - 1) * B + 1, min(n, id * B), tt);
  for (int i = 1; i <= q; i++) {
    vl[Q[i].l].push_back(i);
    vr[Q[i].r].push_back(i);
  }
  tt.clear();
  b[top + 1] = n + 1;
  memset(L, -1, sizeof(L));
  memset(R, -1, sizeof(R));
  for (int i = 1, j = 1; i <= n; i++) {
    for (auto &u : vl[i]) tt.push_back(u);
    if (i == b[j]) {
      for (auto &u : tt) L[u] = j;
      tt.clear();
      j++;
    }
  }
  tt.clear();
  for (int i = n, j = top; i >= 1; i--) {
    for (auto &u : vr[i]) tt.push_back(u);
    if (i == b[j]) {
      for (auto &u : tt) R[u] = j;
      tt.clear();
      j--;
    }
  }
  for (int i = 1; i <= q; i++) {
    if (L[i] == -1 || R[i] == -1 || L[i] > R[i]) continue;
    v[i].push_back(mp[L[i]][R[i]]);
  }
}
int ans[M];
int main() {
  n = read();
  q = read();
  cnt = n;
  for (int i = 1; i <= n; i++) a[i] = read(), pos[a[i]] = i;
  for (int i = 1; i <= q; i++) Q[i].l = read(), Q[i].r = read();
  for (int i = 1; i <= n; i++) bel[i] = (a[i] + B - 1) / B;
  for (int i = 1; i <= (n + B - 1) / B; i++) run(i);
  for (int i = 1; i <= q; i++) {
    ans[i] = v[i][0];
    for (int j = 1; j < (int)(v[i]).size(); j++) {
      ans[i] = Merge(ans[i], v[i][j]);
    }
  }
  cout << cnt << endl;
  for (auto &u : myy) printf("%d %d\n", u.first, u.second);
  for (int i = 1; i <= q; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}