Same Bytes Different Order

4 KB pages, 8-byte doubles, so 512 elements per page. a is page-aligned and spans exactly 8 pages. The TLB is 4-entry, fully associative, LRU, and starts empty. Count only accesses to a.

#define N 4096
double a[N];
double s = 0;

/* loop A */
for (int i = 0; i < N; i++)
    s += a[i];

/* loop B */
for (int j = 0; j < 512; j++)
    for (int i = j; i < N; i += 512)
        s += a[i];

Both loops touch all 4096 elements exactly once. How many TLB misses does each take?