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 N = (1 << 16) + 7;
int cnt, n, Q, B, ind[N], tmp[N], A[N], ans[N], f[43][503][503];
struct qry {
  int l, r, id;
} q1[N], q2[N];
vector<pair<int, int> > op;
bool cmp(int x, int y) { return A[x] < A[y]; }
bool cmpl(qry A, qry B) { return A.l < B.l; }
bool cmpr(qry A, qry B) { return A.r < B.r; }
int merge(int x, int y) {
  if (!x || !y) return (x | y);
  op.push_back(pair<int, int>(x, y));
  return ++cnt;
}
void solve(int id, int L, int R) {
  if (L == R) {
    f[id][1][1] = ind[L];
    return;
  }
  int mid = (L + R) >> 1;
  solve(id + 1, L, mid), solve(id + 2, mid + 1, R);
  for (int i = L; i <= R; ++i) tmp[i] = ind[i];
  inplace_merge(tmp + L, tmp + mid + 1, tmp + R + 1);
  for (int i = L, pl = L, ql = mid + 1; i <= R; ++i) {
    while (pl <= mid && ind[pl] < tmp[i]) ++pl;
    while (ql <= R && ind[ql] < tmp[i]) ++ql;
    for (int j = L, pr = pl - 1, qr = ql - 1; j <= R; ++j) {
      f[id][i - L + 1][j - L + 1] = 0;
      while (pr + 1 <= mid && ind[pr + 1] <= tmp[j]) ++pr;
      while (qr + 1 <= R && ind[qr + 1] <= tmp[j]) ++qr;
      if (pl > pr)
        f[id][i - L + 1][j - L + 1] = f[id + 2][ql - mid][qr - mid];
      else if (ql > qr)
        f[id][i - L + 1][j - L + 1] = f[id + 1][pl - L + 1][pr - L + 1];
      else
        f[id][i - L + 1][j - L + 1] = merge(f[id + 1][pl - L + 1][pr - L + 1],
                                            f[id + 2][ql - mid][qr - mid]);
    }
  }
  for (int i = L; i <= R; ++i) ind[i] = tmp[i];
}
void work(int L, int R) {
  solve(1, L, R);
  static int lft[N], rht[N];
  for (int i = 1, j = L; i <= Q; ++i) {
    while (j <= R && ind[j] < q1[i].l) ++j;
    lft[q1[i].id] = j;
  }
  for (int i = 1, j = L - 1; i <= Q; ++i) {
    while (j + 1 <= R && ind[j + 1] <= q2[i].r) ++j;
    rht[q2[i].id] = j;
  }
  for (int i = 1; i <= Q; ++i)
    if (lft[i] <= rht[i])
      ans[i] = merge(ans[i], f[1][lft[i] - L + 1][rht[i] - L + 1]);
}
int main() {
  cin >> n >> Q;
  for (int i = 1; i <= n; ++i) cin >> A[i], ind[i] = i;
  B = sqrt(Q);
  sort(ind + 1, ind + n + 1, cmp);
  for (int i = 1; i <= Q; ++i) cin >> q1[i].l >> q1[i].r, q1[i].id = i;
  memcpy(q2, q1, sizeof(q1));
  sort(q1 + 1, q1 + Q + 1, cmpl), sort(q2 + 1, q2 + Q + 1, cmpr);
  cnt = n;
  for (int i = 1; i <= n; i += B) work(i, min(n, i + B - 1));
  cout << op.size() + n << endl;
  for (auto t : op) printf("%d %d\n", t.first, t.second);
  for (int i = 1; i <= Q; ++i) printf("%d ", ans[i]);
  return 0;
}