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>
namespace IO {
struct read {
  char buf[10000000], *p1, *p2, c, f;
  read() : p1(buf), p2(buf) {}
  inline char gc(void) {
    return p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, 10000000, stdin),
                       p1 == p2 ? EOF : *p1++)
                    : *p1++;
  }
  inline read &operator>>(int &x) {
    c = gc(), f = 1, x = 0;
    for (; c < '0' || c > '9'; c = gc())
      if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = gc()) x = x * 10 + c - '0';
    x *= f;
    return *this;
  }
};
struct write {
  char buf[10000000], *p1, *p2;
  inline void flush(void) {
    fwrite(buf, 1, p1 - buf, stdout);
    p1 = buf;
  }
  inline void pc(char ch) {
    if (p1 == p2) flush();
    *p1++ = ch;
  }
  write &operator<<(int x) {
    static char b[15];
    static int len(0);
    if (x < 0) pc('-'), x = -x;
    do {
      b[len++] = x % 10 + '0', x /= 10;
    } while (x);
    while (len) {
      len--, pc(b[len]);
    }
    return *this;
  }
  inline write &operator<<(char x) {
    pc(x);
    return *this;
  }
  ~write() { flush(); }
  write() : p1(buf), p2(buf + 10000000) {}
};
read in;
write out;
}  // namespace IO
using IO::in;
using IO::out;
int B, bcnt, n, m, a[4100], mp[4100], bel[4100], ll[4100], rl[4100];
int pool[5000010], *ed = pool;
struct vec {
  int *c, p;
  vec(int const &sz) : c(ed), p(sz) { ed += p * p; }
  int *operator[](int const &x) { return c + x * p; }
  const int *operator[](int const &x) const { return c + x * p; }
};
int cnt;
std::pair<int, int> res[2200010];
typedef std::pair<int *, vec> ret_type;
int merge(int const &a, int const &b) {
  if (!a || !b) return a | b;
  ++cnt;
  res[cnt] = std::make_pair(a, b);
  return cnt;
}
int ans[100010];
ret_type solve(int const &l, int const &r) {
  if (l == r) {
    int *v = ed++;
    *v = mp[l];
    vec vv(1);
    vv[0][0] = mp[l];
    return std::make_pair(v, vv);
  }
  int mid = (l + r) >> 1;
  ret_type ax = solve(l, mid), bx = solve(mid + 1, r);
  int *v = ed;
  int p = ax.second.p + bx.second.p;
  std::merge(ax.first, ax.first + ax.second.p, bx.first, bx.first + bx.second.p,
             v);
  p = (ed = std::unique(v, v + p)) - v;
  vec vv(p);
  for (int i = 0; i < p; i++)
    for (int j = 0; j < p; j++) vv[i][j] = 0;
  int j = 0;
  static int L[100010], R[100010];
  for (int i = 0; i < p; i++) {
    while (j < ax.second.p - 1 && ax.first[j] < v[i]) ++j;
    L[i] = j;
  }
  j = ax.second.p - 1;
  for (int i = p - 1; ~i; i--) {
    while (j && ax.first[j] > v[i]) --j;
    R[i] = j;
  }
  for (int i = 0; i < p; i++)
    if (v[i] <= ax.first[ax.second.p - 1])
      for (int j = i; j < p; j++)
        if (v[j] >= ax.first[0]) vv[i][j] = ax.second[L[i]][R[j]];
  j = 0;
  for (int i = 0; i < p; i++) {
    while (j < bx.second.p - 1 && bx.first[j] < v[i]) ++j;
    L[i] = j;
  }
  j = bx.second.p - 1;
  for (int i = p - 1; ~i; i--) {
    while (j && bx.first[j] > v[i]) --j;
    R[i] = j;
  }
  for (int i = 0; i < p; i++)
    if (v[i] <= bx.first[bx.second.p - 1])
      for (int j = i; j < p; j++)
        if (v[j] >= bx.first[0])
          vv[i][j] = merge(vv[i][j], bx.second[L[i]][R[j]]);
  return std::make_pair(v, vv);
}
struct query {
  int v, id;
  query(int const &vv = 0, int const &i = 0) : v(vv), id(i) {}
} L[100010], R[100010];
struct pp {
  int l, r;
} t[100010];
int main() {
  in >> n >> m;
  B = sqrt(m);
  for (int i = 1; i <= n; i++)
    in >> a[i], mp[a[i]] = i, bel[i] = (i - 1) / B + 1, rl[bel[i]] = i;
  for (int i = 1, l, r; i <= m; i++)
    in >> l >> r, t[i] = (pp){l, r}, ::L[i] = query(l, i), ::R[i] = query(r, i);
  std::sort(L + 1, L + m + 1,
            [](query const &x, query const &y) -> bool { return x.v < y.v; });
  std::sort(R + 1, R + m + 1,
            [](query const &x, query const &y) -> bool { return x.v > y.v; });
  for (int i = n; i; i--) ll[bel[i]] = i;
  bcnt = bel[n];
  cnt = n;
  for (int i = 1; i <= bcnt; i++) {
    ed = pool;
    ret_type bb = solve(ll[i], rl[i]);
    int len = rl[i] - ll[i] + 1;
    int *const &v = bb.first;
    vec const &s = bb.second;
    static int bl[100010], br[100010];
    int k = 0;
    for (int i = 1; i <= m; i++) {
      while (k < len - 1 && L[i].v > v[k]) ++k;
      bl[L[i].id] = k;
    }
    k = len - 1;
    for (int i = 1; i <= m; i++) {
      while (k && R[i].v < v[k]) --k;
      br[R[i].id] = k;
    }
    for (int i = 1; i <= m; i++)
      if (t[i].r >= v[0] && t[i].l <= v[len - 1])
        ans[i] = merge(ans[i], s[bl[i]][br[i]]);
  }
  printf("%d\n", cnt);
  for (int i = n + 1; i <= cnt; i++)
    printf("%d %d\n", res[i].first, res[i].second);
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}