Problem: One fine day Sasha went to the park for a walk. In the park, he saw that his favorite bench is occupied, and he had to sit down on the neighboring one. He sat down and began to listen to the silence. Suddenly, he got a question: what if in different parts of the park, the silence sounds in different ways? So it was. Let's divide the park into 1 × 1 meter squares and call them cells, and numerate rows from 1 to n from up to down, and columns from 1 to m from left to right. And now, every cell can be described with a pair of two integers (x, y), where x — the number of the row, and y — the number of the column. Sasha knows that the level of silence in the cell (i, j) equals to f_{i,j}, and all f_{i,j} form a permutation of numbers from 1 to n ⋅ m. Sasha decided to count, how many are there pleasant segments of silence?

Let's take some segment [l … r]. Denote S as the set of cells (i, j) that l ≤ f_{i,j} ≤ r. Then, the segment of silence [l … r] is pleasant if there is only one simple path between every pair of cells from S (path can't contain cells, which are not in S). In other words, set S should look like a tree on a plain. Sasha has done this task pretty quickly, and called the algorithm — "algorithm of silence's sounds".

Time passed, and the only thing left from the algorithm is a legend. To prove the truthfulness of this story, you have to help Sasha and to find the number of different pleasant segments of silence. Two segments [l_1 … r_1], [l_2 … r_2] are different, if l_1 ≠ l_2 or r_1 ≠ r_2 or both at the same time.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 1000, 1 ≤ n ⋅ m ≤ 2 ⋅ 10^5) — the size of the park.

Each from next n lines contains m integers f_{i,j} (1 ≤ f_{i,j} ≤ n ⋅ m) — the level of silence in the cell with number (i, j).

It is guaranteed, that all f_{i,j} are different.

Output

Print one integer — the number of pleasant segments of silence.

Examples

Input

1 5
1 2 3 4 5


Output

15

Input

2 3
1 2 3
4 5 6


Output

15

Input

4 4
4 3 2 16
1 13 14 15
5 7 8 12
6 11 9 10


Output

50

Note

In the first example, all segments of silence are pleasant.

In the second example, pleasant segments of silence are the following:

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long Inf = 1e18;
const int mod = 1;
template <typename T = int>
T gi() {
  T x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
template <typename T>
bool chkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
};
template <typename T>
bool chkmin(T &a, T b) {
  return a > b ? a = b, 1 : 0;
};
int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
int sub(int a, int b) { return a - b < 0 ? a - b + mod : a - b; }
void inc(int &a, int b) { a = (a + b >= mod ? a + b - mod : a + b); }
void dec(int &a, int b) { a = (a - b < 0 ? a - b + mod : a - b); }
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + "," + to_string(p.second) + ")";
}
template <typename T>
string to_string(T v) {
  int fst = 1;
  string ret = "{";
  for (auto x : v) {
    if (!fst) ret += ",";
    fst = 0, ret += to_string(x);
  }
  ret += "}";
  return ret;
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  dbg_out(T...);
}
template <typename T>
int qpow(int a, T b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % mod;
    a = 1ll * a * a % mod, b >>= 1;
  }
  return ret;
}
struct sgt_data {
  int mn = 0, cnt = 0;
};
struct sgt_tag {
  int val = 0;
};
struct segtree {
  int n = 0;
  vector<sgt_data> tr;
  vector<sgt_tag> tg;
  void reset(int _n) {
    n = _n;
    int N = (n << 2) + 10;
    tr.clear();
    tg.clear();
    tr.resize(N);
    tg.resize(N);
  }
  void apply(int x, int l, int r, int v) {
    tr[x].mn += v;
    tg[x].val += v;
  }
  void pushdown(int x, int l, int r) {
    int mid = (l + r) >> 1;
    if (tg[x].val)
      apply((x << 1), l, mid, tg[x].val),
          apply((x << 1 | 1), mid + 1, r, tg[x].val), tg[x].val = 0;
  }
  sgt_data merge(sgt_data a, sgt_data b) {
    sgt_data ret;
    ret.mn = min(a.mn, b.mn);
    ret.cnt = 0;
    if (ret.mn == a.mn) ret.cnt += a.cnt;
    if (ret.mn == b.mn) ret.cnt += b.cnt;
    return ret;
  }
  void pushup(int x) { tr[x] = merge(tr[(x << 1)], tr[(x << 1 | 1)]); }
  void build(int x, int l, int r) {
    tr[x].cnt = r - l + 1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build((x << 1), l, mid), build((x << 1 | 1), mid + 1, r);
    pushup(x);
  }
  template <typename... T>
  void _modify(int x, int l, int r, int L, int R, T... v) {
    if (L <= l && r <= R) return apply(x, l, r, v...);
    pushdown(x, l, r);
    int mid = (l + r) >> 1;
    if (L <= mid) _modify((x << 1), l, mid, L, R, v...);
    if (R > mid) _modify((x << 1 | 1), mid + 1, r, L, R, v...);
    pushup(x);
  }
  sgt_data _query(int x, int l, int r, int L, int R) {
    if (L <= l && r <= R) return tr[x];
    pushdown(x, l, r);
    int mid = (l + r) >> 1;
    if (R <= mid)
      return _query((x << 1), l, mid, L, R);
    else if (L > mid)
      return _query((x << 1 | 1), mid + 1, r, L, R);
    else
      return merge(_query((x << 1), l, mid, L, R),
                   _query((x << 1 | 1), mid + 1, r, L, R));
  }
  segtree(int n = 0) { reset(n); }
  void build() { build(1, 1, n); }
  template <typename... T>
  void modify(int l, int r, T... v) {
    _modify(1, 1, n, l, r, v...);
  }
  sgt_data query(int l, int r) { return _query(1, 1, n, l, r); }
} sgt;
struct lct_data {
  int sum = 0;
  lct_data() {}
  lct_data(int _sum) { sum = _sum; }
};
struct lct_tag {
  int rev = 0;
  lct_tag() {}
  lct_tag(int _rev) { rev = _rev; }
};
struct LCT {
  vector<vector<int>> ch;
  vector<int> fa;
  vector<lct_data> val, tr;
  vector<lct_tag> tg;
  void reset(int n) {
    const int N = n + 10;
    ch.clear(), fa.clear();
    val.clear();
    tr.clear();
    tg.clear();
    ch.resize(N), fa.resize(N);
    val.resize(N);
    tr.resize(N);
    tg.resize(N);
    for (auto &x : ch) x.resize(2);
  }
  LCT(int n = 0) { reset(n); }
  lct_data merge(lct_data a, lct_data b) {
    lct_data ret;
    ret.sum = a.sum ^ b.sum;
    return ret;
  }
  void apply(int x, bool y = 1, int z = 0) {
    if (y) tg[x].rev ^= 1, swap(ch[x][0], ch[x][1]);
    if (z) val[x].sum ^= z, tr[x].sum ^= z;
  }
  void pushdown(int x) {
    if (tg[x].rev) apply(ch[x][0]), apply(ch[x][1]), tg[x].rev = 0;
  }
  void pushup(int x) {
    tr[x] = merge(merge(tr[ch[x][0]], val[x]), tr[ch[x][1]]);
  }
  bool isroot(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
  void rotate(int x) {
    int y = fa[x], z = fa[y], w = ch[y][1] == x;
    if (!isroot(y)) ch[z][ch[z][1] == y] = x;
    ch[y][w] = ch[x][w ^ 1];
    fa[ch[y][w]] = y;
    ch[x][w ^ 1] = y;
    fa[y] = x;
    fa[x] = z;
    pushup(y);
    pushup(x);
  }
  void pushall(int x) {
    if (!isroot(x)) pushall(fa[x]);
    pushdown(x);
  }
  void splay(int x) {
    pushall(x);
    while (!isroot(x)) {
      int y = fa[x], z = fa[y];
      if (!isroot(y)) (ch[y][0] == x) ^ (ch[z][0] == y) ? rotate(x) : rotate(y);
      rotate(x);
    }
  }
  void access(int x) {
    for (int t = 0; x; t = x, x = fa[x]) splay(x), ch[x][1] = t, pushup(x);
  }
  void makeroot(int x) {
    access(x);
    splay(x);
    apply(x);
  }
  void link(int x, int y) {
    makeroot(x);
    fa[x] = y;
  }
  void cut(int x, int y) {
    makeroot(x);
    access(y);
    splay(y);
    if (ch[y][0] == x && !ch[x][1]) ch[y][0] = fa[x] = 0, pushup(y);
  }
  void split(int x, int y) {
    makeroot(x);
    access(y);
    splay(y);
  }
  int findroot(int x) {
    access(x);
    splay(x);
    while (ch[x][0]) x = ch[x][0];
    splay(x);
    return x;
  }
  template <typename... T>
  void modify(int x, int y, T... v) {
    split(x, y);
    apply(y, v...);
  }
  lct_data query(int x, int y) {
    split(x, y);
    return tr[y];
  }
} lct;
const int N = 1e3 + 10;
int n, m, f[N][N], id[N][N], tot = 0, lp[N * N], dx[4] = {1, 0, 0, -1},
                             dy[4] = {0, 1, -1, 0};
pair<int, int> pos[N * N];
vector<int> get(int i) {
  vector<int> ret;
  int x = pos[i].first, y = pos[i].second;
  for (int d = 0; d < 4; d++) {
    int xx = x + dx[d], yy = y + dy[d];
    if (xx < 1 || xx > n || yy < 1 || yy > m) continue;
    ret.push_back(f[xx][yy]);
  }
  return ret;
}
bool check(int x, int now, int lim, vector<int> v) {
  if (now == int(v.size())) return 1;
  if (v[now] < lim || v[now] > x) return check(x, now + 1, lim, v);
  if (lct.findroot(v[now]) == lct.findroot(x)) return 0;
  lct.link(x, v[now]);
  bool ret = check(x, now + 1, lim, v);
  if (!ret) lct.cut(x, v[now]);
  return ret;
}
int main() {
  n = gi(), m = gi();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) f[i][j] = gi(), id[i][j] = ++tot;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) pos[f[i][j]] = make_pair(i, j);
  lct.reset(n * m), sgt.reset(n * m);
  for (int i = 1, j = 1; i <= n * m; i++) {
    vector<int> v = get(i);
    while (!check(i, 0, j, v)) {
      vector<int> d = get(j);
      for (auto x : d) lct.cut(j, x);
      ++j;
    }
    lp[i] = j;
  }
  sgt.build();
  long long ans = 0;
  for (int i = 1; i <= n * m; i++) {
    sgt.modify(lp[i], i, 1);
    vector<int> v = get(i);
    for (auto x : v)
      if (x >= lp[i] && x <= i) sgt.modify(lp[i], x, -1);
    ans += sgt.query(lp[i], i).cnt;
  }
  printf("%lld\n", ans);
  return 0;
}