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 N = 100010;
const long long M = 320;
const long long INF = 0x3f3f3f3f;
const long long iinf = 1 << 30;
const long long linf = 2e18;
const long long MOD = 1000000007;
const double eps = 1e-7;
void print(long long x) {
  cout << x << endl;
  exit(0);
}
void PRINT(string x) {
  cout << x << endl;
  exit(0);
}
void douout(double x) { printf("%lf\n", x + 0.0000000001); }
long long n, m, len, num;
long long a[N], b[N], bl[N], l[M], r[M], flg[M], flg2[M], sum[M];
void build() {
  len = sqrt(n);
  num = (n - 1) / len + 1;
  for (long long i = (1); i <= (n); i++) bl[i] = (i - 1) / len + 1;
  for (long long i = (1); i <= (num); i++)
    l[i] = (i - 1) * len + 1, r[i] = i * len;
  r[num] = n;
  for (long long i = (1); i <= (num); i++) flg[i] = -1;
}
void down(long long x) {
  for (long long i = (l[x]); i <= (r[x]); i++) a[i] = flg[x];
  flg[x] = -1;
}
void modify(long long x, long long y, long long val) {
  if (bl[x] == bl[y]) {
    if (flg[bl[x]] != -1) down(bl[x]);
    for (long long i = (x); i <= (y); i++) {
      b[i] += abs(a[i] - val);
      sum[bl[x]] += abs(a[i] - val);
      a[i] = val;
    }
  } else {
    if (flg[bl[x]] != -1) down(bl[x]);
    for (long long i = (x); i <= (r[bl[x]]); i++) {
      b[i] += abs(val - a[i]);
      sum[bl[x]] += abs(val - a[i]);
      a[i] = val;
    }
    for (long long i = (bl[x] + 1); i <= (bl[y] - 1); i++) {
      if (flg[i] != -1) {
        flg2[i] += abs(flg[i] - val);
        sum[i] += abs(flg[i] - val) * (r[i] - l[i] + 1);
        flg[i] = val;
      } else {
        for (long long j = (l[i]); j <= (r[i]); j++) {
          b[j] += abs(val - a[j]);
          sum[i] += abs(val - a[j]);
          a[j] = val;
        }
        flg[i] = val;
      }
    }
    if (flg[bl[y]] != -1) down(bl[y]);
    for (long long i = (l[bl[y]]); i <= (y); i++) {
      b[i] += abs(a[i] - val);
      sum[bl[y]] += abs(a[i] - val);
      a[i] = val;
    }
  }
}
long long query(long long x, long long y) {
  long long ans = 0;
  if (bl[x] == bl[y]) {
    for (long long i = (x); i <= (y); i++) ans += b[i] + flg2[bl[i]];
  } else {
    for (long long i = (x); i <= (r[bl[x]]); i++) ans += b[i] + flg2[bl[i]];
    for (long long i = (bl[x] + 1); i <= (bl[y] - 1); i++) ans += sum[i];
    for (long long i = (l[bl[y]]); i <= (y); i++) ans += b[i] + flg2[bl[i]];
  }
  return ans;
}
signed main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = (1); i <= (n); i++) a[i] = i;
  build();
  while (m--) {
    long long op, l, r, val;
    scanf("%lld%lld%lld", &op, &l, &r);
    if (op == 1) {
      scanf("%lld", &val);
      modify(l, r, val);
    } else {
      printf("%lld\n", query(l, r));
    }
  }
  return 0;
}
