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 int INF = 2147483647;
const long long LLINF = 9223372036854775807LL;
const int maxn = 110010;
const int BLOCK = 100;
const int BLCNT = maxn / BLOCK + 10;
long long tcol[maxn];
long long tval[maxn];
long long addsingle[BLCNT];
long long addblock[BLCNT];
long long firstset[BLCNT];
long long tset[BLCNT];
long long tsum[BLCNT];
vector<long long> sorted[BLCNT];
vector<long long> ssum[BLCNT];
long long ABS(long long x) { return x * ((x > 0) - (x < 0)); }
void push(int ind) {
  for (int i = ind * BLOCK; i < (ind + 1) * BLOCK; ++i) {
    tval[i] += addsingle[ind];
    if (firstset[ind] >= 0) tval[i] += ABS(firstset[ind] - tcol[i]);
    if (tset[ind] >= 0) tcol[i] = tset[ind];
  }
  tset[ind] = firstset[ind] = -1;
  addblock[ind] = addsingle[ind] = 0;
}
void rebuild(int ind) {
  sorted[ind].resize(BLOCK);
  tsum[ind] = 0;
  for (int i = 0; i < BLOCK; ++i) {
    sorted[ind][i] = tcol[ind * BLOCK + i];
    tsum[ind] += tval[ind * BLOCK + i];
  }
  sort(sorted[ind].begin(), sorted[ind].end());
  ssum[ind].resize(BLOCK);
  for (int i = 0; i < BLOCK; ++i) {
    ssum[ind][i] = sorted[ind][i];
    if (i) ssum[ind][i] += ssum[ind][i - 1];
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  memset(tset, -1, sizeof(tset));
  memset(firstset, -1, sizeof(firstset));
  while (n % BLOCK) ++n;
  int bcnt = n / BLOCK;
  for (int i = 0; i < n; ++i) tcol[i] = i + 1;
  for (int bb = 0; bb < bcnt; ++bb) {
    rebuild(bb);
  }
  for (int iter = 0; iter < m; ++iter) {
    int qt;
    scanf("%d", &qt);
    if (qt == 1) {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      --l, --r;
      int bl = l / BLOCK;
      int br = r / BLOCK;
      if (bl == br) {
        push(bl);
        for (int i = l; i <= r; ++i) {
          tval[i] += ABS(tcol[i] - x), tcol[i] = x;
        }
        rebuild(bl);
      } else {
        push(bl);
        for (int i = l; i < (bl + 1) * BLOCK; ++i)
          tval[i] += ABS(tcol[i] - x), tcol[i] = x;
        rebuild(bl);
        push(br);
        for (int i = br * BLOCK; i <= r; ++i)
          tval[i] += ABS(tcol[i] - x), tcol[i] = x;
        rebuild(br);
        for (int i = bl + 1; i < br; ++i) {
          if (tset[i] == -1) {
            int pos = lower_bound(sorted[i].begin(), sorted[i].end(),
                                  x - addsingle[i]) -
                      sorted[i].begin();
            --pos;
            long long sum1 = 0;
            int cnt1 = pos + 1;
            if (pos >= 0) sum1 = ssum[i][pos];
            long long sum2 = ssum[i].back() - sum1;
            int cnt2 = BLOCK - cnt1;
            addblock[i] += x * (long long)cnt1 - sum1;
            addblock[i] += sum2 - x * (long long)cnt2;
            firstset[i] = x;
            tset[i] = x;
          } else {
            long long curval = tset[i];
            long long diff = ABS(curval - x);
            addsingle[i] += diff;
            tset[i] = x;
          }
        }
      }
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      --l, --r;
      long long ans = 0;
      int bl = l / BLOCK;
      int br = r / BLOCK;
      if (bl == br) {
        for (int i = l; i <= r; ++i) {
          ans += addsingle[bl];
          ans += tval[i];
          if (firstset[bl] >= 0) ans += ABS(firstset[bl] - tcol[i]);
        }
      } else {
        for (int i = l; i < (bl + 1) * BLOCK; ++i) {
          ans += addsingle[bl];
          ans += tval[i];
          if (firstset[bl] >= 0) ans += ABS(firstset[bl] - tcol[i]);
        }
        for (int i = br * BLOCK; i <= r; ++i) {
          ans += addsingle[br];
          ans += tval[i];
          if (firstset[br] >= 0) ans += ABS(firstset[br] - tcol[i]);
        }
        for (int i = bl + 1; i < br; ++i) {
          ans += tsum[i];
          ans += addsingle[i] * BLOCK;
          ans += addblock[i];
        }
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}
