DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 5e5 + 100;
const long long mod = 1e9 + 7;
const long long inf = 1e18;
long long n, m, seg[maxn], c[maxn];
pair<long long, long long> lz[maxn];
void upd(long long id, long long s, long long e, long long l, long long r,
         long long x) {
  if (r <= s || e <= l) return;
  if (l <= s && e <= r && c[id] != -1) {
    lz[id].first += abs(c[id] - x);
    lz[id].second = x;
    seg[id] += (e - s) * abs(c[id] - x);
    c[id] = x;
    return;
  }
  long long mid = (s + e) / 2;
  if (lz[id].first) {
    seg[id * 2 + 1] += (e - mid) * lz[id].first;
    seg[id * 2] += (mid - s) * lz[id].first;
    c[id * 2] = lz[id].second;
    c[id * 2 + 1] = lz[id].second;
    lz[id * 2].first += lz[id].first;
    lz[id * 2 + 1].first += lz[id].first;
    lz[id * 2].second = lz[id].second;
    lz[id * 2 + 1].second = lz[id].second;
  }
  lz[id].first = 0;
  lz[id].second = -1;
  upd(id * 2, s, mid, l, r, x);
  upd(id * 2 + 1, mid, e, l, r, x);
  seg[id] = seg[id * 2] + seg[id * 2 + 1];
  if (c[id * 2] == c[id * 2 + 1])
    c[id] = c[id * 2];
  else
    c[id] = -1;
  return;
}
long long get(long long id, long long s, long long e, long long l,
              long long r) {
  if (r <= s || e <= l) return 0;
  if (l <= s && e <= r) return seg[id];
  long long mid = (s + e) / 2;
  if (lz[id].first) {
    seg[id * 2 + 1] += (e - mid) * lz[id].first;
    seg[id * 2] += (mid - s) * lz[id].first;
    c[id * 2] = lz[id].second;
    c[id * 2 + 1] = lz[id].second;
    lz[id * 2].first += lz[id].first;
    lz[id * 2 + 1].first += lz[id].first;
    lz[id * 2].second = lz[id].second;
    lz[id * 2 + 1].second = lz[id].second;
  }
  lz[id].first = 0;
  lz[id].second = -1;
  return get(id * 2, s, mid, l, r) + get(id * 2 + 1, mid, e, l, r);
}
int main() {
  long long l, r, x;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  fill(c, c + maxn, 0);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    upd(1, 0, n, i - 1, i, i);
  }
  fill(seg, seg + maxn, 0);
  for (long long i = 0; i < m; i++) {
    cin >> x >> l >> r;
    if (x == 1) {
      cin >> x;
      upd(1, 0, n, l - 1, r, x);
    } else
      cout << get(1, 0, n, l - 1, r) << '\n';
  }
  return 0;
}
